chemistry nya answers part 1

1.2.1 Exercises:

1. Determine the number of significant figures in each of the following

a) 12 2 b) 1098 4 c) 2001 4

d) 2.001 x 103 4 e) 0.000 010 1 3 f) 1.01 x 10-5 3

g) 22.040 30 7 h) 100 1 or 3 i) 100. 3

j) 0.0048 2 k) 0.004 80 3 l) 4.8 x 10-3 2

m) 4.80 x 10-3 3

2. Use the exponential notation to express the number 670 to:

a) Four significant figures 6.700 x 102

b) Three significant figures 6.70 x 102

c) Two significant figures 6.7 x 102

d) One significant figure 7 x 102

1.2.1 Exercises:

3. Perform the following mathematical operations, and express the result to the correct number of significant figures:

a) 97.381 + 4.2502 + 0.099195 = 101.730

b) 21.901 13.21 4.0215 = 4.67

c) 4.0 x 104 x 5.021 x 10-3 x 7.34993 x 102 = 1.5 x 105

d) 2.00 x 106 = 6.67 x 1012

3.00 x 10-7

e) 9.2 x 10.65 = 9.2 x 10.65 = 7.9 8.321 + 4.026 (12.347)

f) (9.04 8.23 + 21.954 + 81.0) 3.1416 = (103.8) = 33.043.1416

1. Convert 15 km/L to miles/gallon, when 1 gallon = 3.7854118 L and 1 km = 0.625 miles.

15 km x 0.625 mile x 3.7854118 l = 35.48823563 miles = 35 milesl km gal gal gal

1.3.1 Exercise:

1. A waterbed has dimensions of 2.55 m x 2.53 dm x 230 cm. What mass of water does this bed contain (density of water = 1.00 g/cm3)?

Volume = L x l x h = 255 cm x 25.3 cm x 230 cm = 1 483 845 cm3 = 1.48 x 106 cm3

Density = m/v m = density x volume = 1 48 x 106 cm3 x 1.00 g/cm3 = 1.48 x 106 g

1.4.1 Exercise:

1.5.1 Exercises:1. Determine the oxidation state of each element in the following compounds:

a) PbCl2 Pb (2+) Cl (1-) b) PbCl4 Pb (4+) Cl (1-)

c) CuCl Cu (1+) Cl (1-) d) Al2O3 Al (3+) O (2-)

e) CrCl3 Cr (3+) Cl (1-)

1. Supply the following names or formulas:

a) K2O potassium oxide c) Zn3N2 zinc nitride

c) BaS barium sulfide d) AlH3 aluminum hydride

e) Silver oxide Ag2O f) Beryllium iodide BeI2

g) Lithium phosphide Li3P h) Calcium carbide Ca2C

1.5.2 Exercises:

1. Name the following compounds:

a) CuCl copper(I) chloride b) HgO mercury (II) oxide c) Fe2O3 iron(III) oxide

d) MnO2 manganese(IV) oxide e) PbCl2 lead(II) chloride f) CoBr2 cobalt(II) bromide

g) CaCl2 calcium chloride h) Al2O3 aluminum oxide i) CrCl3 chromium(III) chloride

1.5.3 Exercises:

1.5.4 Exercises:


a) KMnO4 potassium permanganate b) Ba(OH)2 barium hydroxide

c) Fe(OH)3 iron(III) hydroxide d) NaH2PO4 sodium dihydrogen phosphate

e) (NH4)2Cr2O7 ammonium dichromate f) NaBrO4 sodium perbromate

g) KIO3 potassium iodate h) Ca(OCl)2 calcium hypochlorite

i) Cr(NO3)3 chromium(III) nitrate

1. Name the following molecules:

a) N2O dinitrogen monoxide b) NO nitrogen monoxide

c) NO2 nitrogen dioxide d) N2O3 dinitrogen trioxide

e) N2O4 dinitrogen tetroxide g) N2O5 dinitrogen pentoxide

1.5.5 Exercises:

1.5.6 Exercises:


a) HCl (aq) hydrochloric acid b) HNO3 nitric acid

c) HNO2 nitrous acid d) HF (aq) hydrofluoric acid

e) HBr (aq) hydrobromic acid f) H2SO4 sulfuric acid

g) H2SO3 sulfurous acid h) HI (aq) hydroiodic acid

i) HCN (aq) hydrocyanic acid j) H3PO4 phosphoric acid

k) H2S (aq) hydrosulfuric acid l) HC2H3O2 acetic acid

1. Name the following:

a) CuSO45H2O copper(II) sulfate pentahydrate

b) Na2CO310H2O sodium carbonate decahydrate

c) MgSO47H2O magnesium sulfate heptahydrate

1.5.7 Exercises:

1.5.8 Exercises:

N2O3 dinitrogen trioxide Mg(OH)2 magnesium hydroxide

(NH4)2CO3 ammonium carbonate Fe3(PO4)2 iron(II) phosphate

CoBr3 cobalt(III) bromide CaI2 calcium iodide

PCl5 phosphorous pentachloride P2O5 diphosphorous pentoxide

FeCl2 iron(II) chloride NaNO3 sodium nitrate

Na2CrO4 sodium chromate HBrO hypobromous acid

CuSO3 copper(II) sulfite Ca3N2 calcium nitride

Li2SO4 lithium sulfate HgS mercury(II) sulfide

KNO2 potassium nitrite LiH lithium hydride

FeCr2O7 iron(II) dichromate N2O dinitrogen monoxide

NaHCO3 sodium hydrogen carbonate Ba(NO3)2 barium nitrate

KHSO4 potassium hydrogen sulfate Ba(ClO3)2 barium chlorate

Cu2O copper(I) oxide AlPO4 aluminum phosphate

Ba(C2H3O2)2 barium acetate NaIO3 sodium iodate

Cu(CN)2 copper(II) cyanide K2SO3 potassium sulfite

SO3 sulfur trioxide Mg(ClO4)2 magnesium perchlorate

H3PO4 phosphoric acid (NH4)2SO4 ammonium sulfate

N2O5 dinitrogen pentoxide NaC2H3O2 sodium acetate

HNO2 nitrous acid KMnO4 potassium permanganate


Magnesium sulfide MgS Tin (II) bromide SnBr2Calcium hydroxide Ca(OH)2 Lithium hydrogen carbonate LiHCO3Hydroiodic acid HIO3 Calcium fluoride CaF2Aluminum nitrate Al(NO3)3 Ammonium hydrogen sulfate NH4HSO4Copper (I) cyanide CuCN Sodium sulfide Na2SIron (II) sulfate FeSO4 Phosphorous acid H3PO3Perbromic acid HBrO4 Carbon monoxide COIron (III) bromide FeBr3 Potassium chromate K2CrO4Zinc permanganate Zn(MnO4)2 Hydrogen peroxide H2O2Potassium nitrate KNO3 Calcium nitride Ca3N2Iron (III) chromate Fe2(CrO4)3 Dinitrogen trioxide N2O3Sulfurous acid H2SO3 Magnesium chloride MgCl2Magnesium chlorate Mg(ClO3)2 Ammonium carbonate (NH4)2CO3Cobalt (II) hydroxide Co(OH)2 Nitric acid HNO3Sodium hydroxide NaOH Diphosphorus pentasulfide P2S5Lithium sulfite Li2SO3 Sodium hypochlorite NaClOAmmonium dichromate (NH4)2Cr2O7 Mercury (II) nitrate Hg(NO3)2Diphosphorus pentoxide P2O5 Sulfuric acid H2SO4Iron (III) carbonate Fe2(CO3)3 Calcium acetate Ca(C2H3O2)2Nickel (II) perchlorate Ni(ClO4)2 Potassium sulfite K2SO3Magnesium hydrogen sulfite Mg(HSO3)2 Tin (IV) bromide SnBr4Sulfur hexafluoride SF6 Iron (III) cyanide Fe(CN)3Ammonium carbonate (NH4)2CO3 Phosphoric acid H3PO4Lithium dichromate Li2Cr2O7 Iron (II) hydroxide Fe(OH)2

1.5.8 Exercises:2. Write the formulas for the following compounds:

1.6.1 Exercises:

1. Calculate the molar mass of calcium carbonate.Ca(CO3)2: M = Ca + 2C + 6O = 40.08 g/mol + 2 x 12.01 g/mol + 6 x 16.00 g/mol = 160.10 g/mol

2. A certain sample of calcium carbonate contains 4.86 moles. What is the mass (in grams) of the sample? What is the mass of carbonate ions present?n = m/M m = nM = 4.86 mol x 160.10 g/mol = 778.086 g = 778 g

3. Convert 4.80 g of KHC8H4O4 to the number of moles of KHC8H4O4.n = m/M = 4.80 g / 204.23 g/mol = 0.023503 mol = 2.35 x 10-2 mol

4. Convert 0.01132 mol of MgSO47H2O to its number of grams.n = m/M m = nM = 0.01132 mol x 246.52 g/mol = 2.790606 g = 2.791 g

5. What is the mass of water in 0.01132 mol of MgSO47H2O?m = nM = 0.01132 mol x 18.02 g/mol = 0.2039864 g = 0.2040 g

6. Convert 362 g of H2SO4 to the number of molecules of H2SO4.n = m/M = 362 g / 98.09 g/mol = 3.6904883 mol = 3.69 mol

3.69 mol x 6.022 x 1023 molecules = 22.22118 x 1023 molecules = 2.22 x 1024 molecules.1 mol

1.6.1 Exercises:7. Convert 9.0 x 1015 molecules of KHC8H4O4 to the number of moles of KHC8H4O4.9.0 x 1015 molecules x 1 mole/6.022 x 1023 molecules = 1.4 x 10-8 moles

8. For aspartame, C14H18N2O5a) How many hydrogen atoms are present in 10.0 g of aspartame?

n = m/M = 10.0 g / 294.34 g/mol = 0.033974315 mol = 0.0340 mol of aspartame0.0340 moles of aspartame x 18 moles of hydrogen / mole of aspartame = 0.612 moles of hydrogen0.612 moles of hydrogen x 6.022 x 1023 atoms / mole of hydrogen = 3.685464 1023 atoms of hydrogen

= 3.69 x 1023 atoms of hydrogen

b) What is the mass in grams of one molecule of aspartame?294.34 g/mol x 1 mole / 6.022 x 1023 molecules = 4.887745 x 10-22 g/molecule

= 4.888 x 10-22 g/molecule

9. Isopentyl acetate (C7H14O2) is a compound responsible for the scent of bananas. Interestingly, bees release about 1 g of this compound when they sting.

a) How many molecules of isopentyl acetate are released in a typical bee sting?n = m/M = 1 x 10-6 g / 130.21 g/mol = 7679902 x 10-9 mol = 8 x 10-9 mol8 x 10-9 mol x 6.022 x 1023 molecules / 1 mol = 4.8176 x 1015 molecules = 5 x 1015 molecules

b) How many carbon atoms are present?5 x 1015 molecules x 7 carbons / 1 molecule = 35 x 1015 carbons = 4 x 1016 carbons

1.7.1 Exercises:1. Calculate the mass percent (with 4 s.f.) of each element in:

a) Carvone, C10H14O M = 150.24 g/mol%C = 10 x 12.01 g/mol x 100% = 79.93876% = 79.94%

150.24 g/mol%H = 14 x 1.01 g/mol x 100% = 9.4116081% = 9.412%

150.24 g/mol%O = 16.00 g/mol x 100% = 1.064963% = 1.065%

150.24 g/mol

b) Ca(H2PO4)2 M = 234.06 g/mol%Ca = 40.08 g/mol x 100% = 17.12%

234.06 g/mol%H = 4 x 1.01 g/mol x 100% = 1.726%

234.08 g/mol%P = 2 x 30.97 g/mol x 100% = 26.46%

234.08 g/mol%O = 8 x 16.00 g/mol x 100% = 54.68%

234.08 g/mol

1.7.1 Exercises (continuation):

c) Penicillin F, C14H20N2SO4 M = 312.43 g/mol%C = 14 x 12.01 g/mol x 100% = 53.82%

312.43 g/mol%H = 20 x 1.01 g/mol x 100% = 6.465 %

312.43 g/mol%N = 2 x 14.01 g/mol x 100% = 8.968%

312.43 g/mol%S = 32.07 g/mol x 100% = 10.26%

312.43 g/mol%O = 4 x 16.00 g/mol x 100% = 20.49%

312.43 g/mol

1.8.1 Exercises:1. Analysis of the hydrocarbon CxHy reveals a hydrogen content of 11.84% by mass.

What is the empirical formula?For H: n = m/M = 11.84 g/1.01 g/mol = 11.722772 mol = 11.72 molFor C: n = m/M = 88.16 g/12.01 g/mol = 7.305495 mol = 7.306 molRatio (H/C): 11.72 mol / 7.306 mol = 1.6For every C there is 1.6 H. For every 5 carbons, there is 8 hydrogens: C5H8

2. Consider the following reaction: Li2CO3 (s) + Co2O3 (s) CO2 (g) + black compoundThe black compound contains 7.09% Li, 60.2% Co and 32.7% O. Identify the empirical formula.

For Li: n = m/M = 7.09 g/6.94 g/mol = 1.02 molFor Co: n = m/M = 60.2 g/58.93 g/mol = 1.02 molFor O: n = m/M = 32.7 g/16.00 g/mol = 2.04 mol LiCoO2

3. A caffeine sample contains 0.624 g C, 0.065 g H, 0.364 g N and 0.208 g O. What is the empirical formula of caffeine (molecular weight of = 194.2 g/mol)? What is the molecular formula of caffeine?C: n = m/M = 0.624 g/12.01 g/mol = 5.196 x 10-2 mol H: n = m/M = 0.065 g/1.01 g/mol = 6.436 x 10-2 molN: n = m/M = 0.364 g/14.01 g/mol = 2.598 x 10-2 mol O: n = m/M = 0.208 g/16.00 g/mol = 1.300 x 10-2 molC4H5N2O since the empirical formula affords a MW of 97.07 g/mol, and since the molecular weight is double (194.2 g/mol), the empirical formula must be multiplied by 2 to afford C8H10N4O2.


4. The antibiotic sulphanilamide has the following percent composition: 41.9% C, 4.65% H, 18.6% S, 16.3% N and 18.6% O.Find the empirical formula of sulphanilamide. The molecular weight of sulphanilamide is 172 g. What is the molecular formula?

ratio:C: n = m/M = 41.9 g/12.01 g/mol = 3.49 mol 6H: n = m/M = 4.65 g/1.01 g/mol = 4.60 mol 8S: n = m/M = 18.6 g/32.07 g/mol = 0.580 mol 1N: n = m/M = 16.3 g/14.01 g/mol = 1.16 mol 2O: n = m/M = 16.8 g/16.00 g/mol = 1.05 mol 2

C6H8SN2O2 empirical formula: 172.23 g/mol

The molecular formula is the same as the empirical formula.

1.9.1 Exercises:

1. Balance the following equations

a) CH4 (g) + O2 (g) CO2 (g) + H2O (l)

b) (NH4)2Cr2O7 (s) Cr2O3 (s) + N2 (g) + H2O (l)

c) NH3 (g) + O2 (g) NO (g) + H2O (l)

d) Ca3(PO4)2 (s) + H2SO4 (aq) CaSO4 (s) + H3PO4 (aq)

e) NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + H2O (l)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

(NH4)2Cr2O7 (s) Cr2O3 (s) + N2 (g) + 4 H2O (l)

2 NH3 (g) + 2 O2 (g) 2 NO (g) + 3 H2O (l)

Ca3(PO4)2 (s) + 3 H2SO4 (aq) 3 CaSO4 (s) + 2 H3PO4 (aq)

3 NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3 H2O (l)

Exercises:1. In the combustion of propane (C3H8), what mass of oxygen will react with 96.1 g of propane?

C3H8 (g) + O2 (g) CO2 (g) + H2O (l)C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

n = m/M = 96.1 g/44.11 g/mol = 2.18 mol of propane2.18 mol of propane x 5 mol of O2/1 mol of propane = 10.9 mol of O2m = nM = 10.9 mol x 32.00 g/mol = 349 g of O2.

2. Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess HCl in your stomach:NaHCO3 (aq) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g)

Milk of magnesia (Mg(OH)2), is also used as an antacid:Mg(OH)2 (s) + HCl (aq) 2 H2O (l) + MgCl2 (aq)

Which is more effective per 1.00 g of antacid?NaHCO3 (aq) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g)

n = m/M = 1.00 g/84.01 g/mol = 1.19 x 10-2 mol NaHCO31.19 x 10-2 mol NaHCO3 x 1 mol HCl/1 mol NaHCO3 = 1.19 x 10-2 mol HCl

Mg(OH)2 (s) + 2 HCl (aq) 2 H2O (l) + MgCl2 (aq)n = m/M = 1.00 g/58.33 g/mol = 1.71 x 10-2 mol Mg(OH)21.71 x 10-2 mol Mg(OH)2 x 2 mol HCl/1 mol Mg(OH)2 = 3.42 x 10-2 mol HCl

Mg(OH)2 is more effective.

1.11.1 Exercises:1. Consider the following reaction: HNO3 (aq) + Ca(OH)2 (s) Ca(NO3)2 (aq) + H2O (l)

a) What weight of calcium nitrate can be produced by the reaction of 18.9 g of nitric acid with 7.4 g of calcium hydroxide? b) How many moles of which reagent will remain unreacted?

2 HNO3 (aq) + Ca(OH)2 (s) Ca(NO3)2 (aq) + 2 H2O (l)a) n = m/M = 18.9 g/63.02 g/mol = 3.00 x 10-1 mol HNO3

n = m/M = 7.4 g/74.10 g/mol = 1.0 x 10-2 mol Ca(OH)23.00 x 10-1 mol HNO3 x 0.5 mol Ca(OH)2/mol HNO3 = 1.50 x 10-1 mol Ca(OH)2, Limiting reagent: Ca(OH)21.0 x 10-2 mol Ca(OH)2 x 1 mol Ca(NO3)2/1 mol Ca(OH)2 = 1.0 x 10-2 mol Ca(NO3)2 m = nM = 1.0 x 10-2 mol x 164.10 g/mol = 1.64 g Ca(NO3)2b) 1.0 x 10-2 mol Ca(OH)2 x 2 mol HNO3/1 mol Ca(OH)2 = 2 x 10-2 mol of Ca(OH)2What is left over: 3.00 x 10-2 mol 1.0 x 10-2 mol = 2.0 x 10-2 mol HNO3

2. Consider the following reaction: P (s) + S (s) P4S3 (s)a) How many grams of tetraphosphorus trisulfide can be produced from 62.0 g of phosphorus and

38.6 g of sulfur? B) How many grams of which reagent will remain unreacted?4 P (s) + 3 S (s) P4S3 (s)

a) n = m/M = 62.0 g/30.97 g/mol = 2.00 mol Pn = m/M = 38.6 g/32.07 g/mol = 1.20 mol S2.00 mol P x 3 mol S/4 mol P = 1.50 mol S: The limiting reagent is S.1.20 mol S x 1 mol P4S3/3 mol S = 0.400 mol P4S8m = nM = 0.400 mol x 220.06 g/mol = 88.0 g of P4S3b) Excess reagent: 1.2 mol S x 4 mol P/3 mol S = 1.6 mol P will be used upUnreacted reagent: 2.00 mol 1.6 mol = 0.400 mol of P will remain.M = nM = 0.400 mol x 30.97 g/mol = 12.388 g of P will remain.

1.11.1 Exercises:

3. Consider the following reaction: HBr (aq) + Fe(OH)3 (s) FeBr3 (aq) + H2O (l)a) What weight of iron (III) bromide can be produced by the reaction of 30.0 g of hydrogen bromide with 10.0 g of iron (III) hydroxide? B) Given that the percentage yield is 66%, how many grams of iron (III) bromide are obtained?

3 HBr (aq) + Fe(OH)3 (s) FeBr3 (aq) + 3 H2O (l)a) HBr: n = m/M = 30.0 g/80.91 g/mol = 0.371 mol HBrFe(OH)3: n = m/M = 10.0 g/106.88 g/mol = 0.0936 mol Fe(OH)3Limiting reagent calculations: 0.371 mol HBr x 1 mol Fe(OH)3/3 mol HBr = 0.124 mol Fe(OH)3Fe(OH)3 is the limiting reagent.0.0936 mol Fe(OH)3 x 1 mol FeBr3/1 mol Fe(OH)3 = 0.0936 molm = nM = 0.0936 mol x 295.55 g/mol = 27.66 g FeBr3b) 66% x 27.66 g = 18.26 g

1.12.1 Exercises:1. A solution can be prepared by mixing 1.00 g of ethanol (C2H5OH) with 100.0 g of water to give a final volume of 101 mL. Calculate the molarity of ethanol in the solution.n = m/M = 1.00 g/46.08 g/mol = 0.0217 molM = n/v = 0.0217 mol/0.101 l = 0.215 mol/L

2. A sulphuric acid solution of density 1.802 g/mL contains 88.0% H2SO4 by weight. A) Find the weight of H2SO4 per liter of solution. B) Find the molarity of the solution.a) 1 liter = 1 802 g 0.880 x 1 802 g = 1.59 x 103 gb) n = m/M = 1.59 x 103 g/98.09 g/mol = 16.2 molM = n/v = 16.2 mol/1 L = 16.2 mol/L

3. A 325 mL sample of a solution contains 25.3 g CaCl2 (111 g/mol). A) Calculate the molarity of Cl-

ions in solution. B) How many grams of Cl- are there in 0.100L of this solution?a) n = m/M = 25.3 g/110.98 g/mol = 0.2280 molM = n/v = 0.2280 mol/0.325 L = 0.702 mol/Lb) n = MV = 0.702 mol/L x 0.100 L = 0.0702 molm = nM = 0.0702 mol x 35.45 g/mol = 2.49 g of Cl-.

4. Find the weight percent of calcium carbonate in a 2.00 g sample of rock if it reacts with 30.0 mL of 0.60 M H2SO4 according to the following reaction: CaCO3 (s) + H2SO4 (aq) CaSO4 (s) + H2O (l) + CO2 (g)n = MV = 0.60 mol/L x 0.0300 L = 0.0180 molm = nM = 0.0180 mol x 100.09 g/mol = 1.80 gWeight % = 1.80 g/2.00 g x 100% = 90.0%

5. What volume of a 0.200 M solution of AgNO3 is required to react with 12.2 g of K3PO4?AgNO3 (aq) + K3PO4 (aq) Ag3PO4 (s) + KNO3 (aq)

3 AgNO3 (aq) + K3PO4 (aq) Ag3PO4 (s) + 3 KNO3 (aq)n = m/M = 12.2 g/212.27 g/mol = 0.0575 molM = n/v v = n/M = 0.0575 mol/0.200 mol/L = 0.288 L

6. What mass of Fe(OH)3 would be produced by reacting 75.00 mL of 0.105 M Fe(NO3)3 with 125.0 mL of 0.150 M NaOH? NaOH (aq) + Fe(NO3)3 (aq) Fe(OH)3 (s) + NaNO3 (aq)

3 NaOH (aq) + Fe(NO3)3 (aq) Fe(OH)3 (s) + 3 NaNO3 (aq)NaOH: n = MV = 0.125.0 L x 0.150 mol/L = 0.0188 molFe(NO3)3: n = MV = 0.07500 L x 0.105 mol/L = 0.00788 molLimiting reagent: 0.0188 mol NaOH x 1 mol Fe(NO3)3/3 mol NaOH = 0.00627 mol Fe(NO3)3NaOH is the limiting reagent.Mole ratio: 0.0188 mol NaOH x 1 mol Fe(OH)3/3 mol NaOH = 0.00627 mol Fe(OH)3m = nM = 0.00627 mol x 106.88 g/mol = 0.670 g of Fe(OH)3.


7. The percentage of sodium chloride in a mixture was determined by adding 0.0811 M AgNO3 to a 0.300 g sample of the mixture dissolved in water. It was found that 27.30 mL of AgNO3 solution was required for a complete reaction: NaCl (aq) + AgNO3 (aq) AgCl (s) + NaNO3 (aq)a) What was the percent by mass of sodium chloride in the mixture? B) How many grams of AgClprecipitated? C) Given that the final volume of the solution is 50.00 mL, calculate the molarity of sodium nitrate in that solution.a) AgNO3: n = Mv = 0.0811 mol/L x 0.02730 L = 0.002214 mol AgNO3

NaCl: 0.002214 mol AgNO3 x 1 mol NaCl/1 mol AgNO3 = 0.002214 mol NaClm = nM = 0.002214 mol x 58.44 g/mol = 0.1294 g

Mass % = 0.1294 g/0.300 g x 100% = 43.13 %b) m = nM = 0.002214 mol x 169.88 g/mol = 0.3761 g of AgCl.c) NaNO3: 0.002214 mol

M = n/v = 0.002214 mol/0.05000 L = 0.04428 mol/L


8. A sample of magnesium metal reacted completely when allowed to stand in 0.100 L of 0.1000 M HCl. The excess acid was neutralized by titration with 32.0 mL of 0.125 M KOH. What was the mass of the sample of magnesium? Mg (s) + HCl (aq) MgCl2 (aq) + H2 (g)

HCl (aq) + KOH (aq) KCl (aq) + H2O (l)Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)HCl (aq) + KOH (aq) KCl (aq) + H2O (l)

HCl: n = Mv = 0.1000 mol/L x 0.100 L = 0.0100 mol of HClTitration: n = Mv = 0.125 mol/L x 0.0320 L = 0.00400 molThe amount that has reacted is the difference between the initial amount of HCl used & the left over:0.0100 mol 0.00400 mol = 0.00600 mol (0.00600 mol of HCl was reacted)Mole ratio: 0.00600 mol HCl x 1 mol Mg/2 mol HCl = 0.00300 mol Mgm = nM = 0.00300 mol x 24.31 g/mol = 0.0729 g of Mg

9. Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate. How many millilitres of 0.125 M NaHCO3 solution are needed to neutralize 18.0 mL of 0.100 M HCl? HCl (aq) + NaHCO3 (aq) NaCl (aq) + H2O (l) + CO2 (g)HCl: n = Mv = 0.100 mol/L x 0.0180 L = 0.00180 molNaHCO3: v = n/M = 0.00180 mol/0.125 mol/L = 0.0144 L = 144 mL are necessary.

10. What volume of 0.250 M H2SO4 is needed to react with 50.0 mL of 0.100 M NaOH?H2SO4 (aq) + NaOH (aq) Na2SO4 (aq) + H2O (l)

H2SO4 (aq) + 2 NaOH (aq) Na2SO4 (aq) + 2 H2O (l)NaOH: n = Mv = 0.100 mol/L x 0.0500 L = 0.00500 molMole ratio: 0.00500 mol NaOH x 1 mol H2SO4/2 mol NaOH = 0.00250 mol H2SO4H2SO4: v = n/M = 0.00250 mol/0.250 mol/L = 0.0100 L of H2SO4.

1.12.1. Exercises:

1.13.1 Exercises:

1. You have 505 mL of a 0.125 M HCl solution and you want to dilute it to 0.100 M. How much water should you add?

M1 = 0.125 mol/L V1 = 0.505 L M2 = 0.100 mol/L V2 = ?M1V1 = M2V2 V2 = M1V1/M2 = 0.125 mol/L x 0.505 L/0.100 mol/L = 0.631 LWater needed: 0.631 L 0.505 L = 0.236 L

2. Describe how to prepare 1.500L of 0.8460 M HBr solution starting with a 2.000 M solution.M1 = 2.000 mol/L V1 = ? M2 = 0.8460 mol/L V2 = 1.500 LM1V1 = M2V2 V1 = M2V2/M1 = 0.8460 mol/L x 1.500 L/2.000 mol/L = 0.6345 L

0.6345 L of the 2.000 mol/L HBr solution needs to be taken and added to a 1.500 L volumetric flask , and then water is added until the final volume is 1.500 L.

1.14.1 Exercises:1. Consider the following reaction: CH3OH (l) + O2 (g) CO2 (g) + H2O (g)If 40.0 g of methanol (CH3OH) is placed in a reaction vessel with 46.0 g of oxygen gas, a total of 19.3 liters of CO2 is actually obtained at STP. A) What is the % yield of CO2? B) How many grams of CO2were formed? CH3OH (l) + 1 O2 (g) CO2 (g) + 2 H2O (g)a) CH3OH: n = m/M = 40.0 g/32.05 g/mol = 1.25 mol

O2: n = m/M = 46.o g/32.00 g/mol = 1.44 molLimiting reagent: 1.25 mol CH3OH x 1 mol O2/1 mol CH3OH = 1.88 mol O2O2 is the limiting reagent.Mole ratio: 1.44 mol O2 x 1 mol CO2/1 mol O2 = 0.960 mol of CO2 (theor.)

pv = nRT n = pv/RT = 1 atm x 19.3 L/0.08206 L atm/mol K x 273 K = 0.862 mol (exp.)%yield = 0.862 mol/0.960 mol x 100% = 89.8%b) m = nM = 0.862 mol x 44.01 g/mol = 27.94 g

2. Consider the following reaction: CCl4 (l) + SbF3 (l) CCl2F2 (g) + SbCl3 (l)a) If 150 g of CCl4 are mixed with 200 g of SbF3, how many grams of CCl2F2 can be formed?b) Calculate the volume of CCl2F2 gas at 30 C and 0.90 atm.

3 CCl4 (l) + 2 SbF3 (l) 3 CCl2F2 (g) + 2 SbCl3 (l)a) CCl4: n = m/M = 150 g/153.81 g/mol = 0.975 mol

SbF3: n = m/M = 200 g/178.76 g/mol = 1.12 molLimiting reagent: 0.975 mol CCl4 x 2 mol SbF3/3 mol CCl4 = 0.650 mol SbF3The limiting reagent is CCl4.Mole ratio: 0.975 mol CCl4 x 3 mol CCl2F2/3 mol CCl4 = 0.975 mol CCl2F2

m = nM = 0.975 mol x 120.91 g/mol = 118 g CCl2F2b) pv = nRT v = nRT/p = 0.975 mol x 0.08206 L atm/mol K x 303 K/0.90 atm = 27 L of CCl2F3

1.14.1 Exercises:3. How many liters of gaseous hydrogen at 150 C and 1.34 atm will be produced when 55.0 g of aluminum metal are reacted with 425 mL of 6.00 M sulphuric acid?

Al (s) + H2SO4 (aq) Al2(SO4)3 (aq) + H2 (g)2 Al (s) + 3 H2SO4 (aq) Al2(SO4)3 (aq) + 3 H2 (g)

Al: n = m/M = 55.0 g/26.98 g/mol = 2.04 molH2SO4: n = Mv = 6.00 mol/L x 0.425 L = 2.55 molLimiting reagent: 2.04 mol Al x 3 mol H2SO4/2 mol Al = 3.06 mol H2SO4The limiting reagent is H2SO4.Mole ratio: 2.55 mol H2SO4 x 3 mol H2/3 mol H2SO4 = 2.55 mol H2

pv = nRT v = nRT/p = 2.55 mol x 0.08206 L atm/mol K x 423 K/1.34 atm = 66.1 L

4. Consider the following reaction: MnO2 (s) + HCl (aq) MnCl2 + Cl2 (g) + H2O (l)a) Calculate the expected volume of chlorine gas, measured at 812 mm Hg and 24 C, when 1.44 g of MnO2 reacted with 50.0 mL of 1.25 M HCl solution. b) If 278 mL of chlorine gas was collected at the same temperature and pressure, find the percentage yield.

MnO2 (s) + 4 HCl (aq) MnCl2 + Cl2 (g) + 2 H2O (l)a) MnO2: n = m/M = 1.44 g/86.94 g/mol = 0.0166 mol

HCl: n = Mv = 1.25 mol/L x 0.0500 L = 0.0625 molLimiting reagent: 0.0166 mol MnO2 x 4 mol HCl/1 mol MnO2 = 0.0664 mol HClThe limiting reagent is HCl.Mole ratio: 0.0625 mol HCl x 1 mol Cl2/4 mol HCl = 0.0156 molpv = nRT v = nRT/p = 0.0156 mol x 0.08206 L atm/mol K x 297 K/1.07 atm = 0.355 Lb) % yield = 0.278 L/0.355 L x 100% = 78.3%

5. A sample of diborane gas, B2H6, has a pressure of 345 mm Hg at a temperature of -15 C and a volume of 3.48 L. If conditions were changed so that the temperature is 36 C and the pressure is 468 mm Hg, what will be the volume of this sample?P1 = 0.454 atm T1 = 258 K V1 = 3.48 LP2 = 0.616 atm T2 = 310 K V2 = ?P1V1/T1 = P2V2/T2 V2 = P1V1T2/T1P2 = 0.454 atm x 3.48 L x 310 K/ (258 K x 0.616 atm) = 3.08 L

6. Hydrogen gas is often produced by the reaction of magnesium and hydrochloric acid:Mg (s) + HCl (aq) MgCl2 (aq) + H2 (g)

If sufficient acid were used to react completely with 7.000 g Mg, what volume of hydrogen would be collected on a day when the temperature was 28 C and the pressure was 765 mm Hg?

Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)Mg: n = m/M = 7.000 g/24.31 g/mol = 0.2880 molMole ratio: 0.2880 mol Mg x 1 mol H2/1 mol Mg = 0.2880 mol H2pv = nRT v = nRT/p = 0.2880 mol x 0.08206 L atm/mol K x 301 K/1.01 atm = 7.04 L

1.14.1 Exercises:

1.15.1 Exercises:1. Mixtures of helium and oxygen are used in scuba diving tanks to help prevent the bends. For a particular dive, 46 L of O2 at 25 C and 1.0 atm, and 12 L of He at 25 C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25 C.O2: V1 = 46 L T1 = 298 K P1 = 1.o atm

V2 = 5.0 L T2 = 298 K P2 = ?P1V1 = P2V2 P2 = P1V1/V2 = 1.0 atm x 46 L/5.0 L = 9.2 atm PO2 = 9.2 atm

He: V1 = 12 L T1 = 298 K P1 = 1.0 tmV2 = 5.0 L T2 = 298 K P2 = ?

P1V1 = P2V2 P2 = P1V1/V2 = 1.0 atm x 12 L/5.0 L = 2.4 atm PHe = 2.4 atm

Ptotal = PO2 + PHe = 9.2 atm + 2.4 atm = 11.6 atm

2. If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in an evacuated 1.04 L contained at 25 C, what will be the partial pressure of each gas and the total pressure in the container?

He: n = m/M = 2.41 g/4.00 g/mol = 0.603 molpv = nRT p = nRT/v = 0.603 mol x 0.08206 L atm/mol K x 298 K/1.04 L = 14.2 atm

PHe = 14.2 atm

Ne: n = m/M = 2.79 g/20.18 g/mo = 0.138 molpv = nRT p = nRT/v = 0.138 mol x 0.08206 L atm/mol K x 298 K/1.04 L = 3.25 atm

PNe = 3.25 atm

Ptotal = PHe + PNe = 14.2 atm + 3.25 atm = 17.45 atm

chemistry nya answers part 1

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