unit 8 test review1.doc

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Name_______________________________

Period_____

Unit 8: Stoichiometry Test Review

Indicate whether the statement is true or false. (1-11)

____ 1. Stoichiometryis the study of the relationship between the amount of reactants used and the amount of

products made in a chemical reaction.

____ 2. In a balanced chemical equation, the total number of moles of all the reactants is equal to the total number ofmoles of all the products.

____ 3. In a chemical reaction represented by the general equation , there are four distinct mole ratiosthat can be written.

____ 4. The mole ratiois a comparison of how many moles of one substance are required to participate in a chemical

reaction with another substance, based on the balanced chemical equation.

____ . In a chemical reaction, the reactant with the largest molar mass is the limiting reagent.

____ !. The excess reagentrefers to the additional amount of a reactant in a chemical reaction that must be added in

order for the reaction to proceed.

____ ". The stoichiometric relationship between any two substances in a reaction depends on the mole ratio between

those substances.

____ #. The limiting reagentlimits the amount of product formed in a chemical reaction.

____ $. In a balanced chemical equation, the total number mass of the reactants is equal to the total mass of the

products.

____ 1%. The percent yield is the ma&imum amount of product that can be produced from a gi'en amount of reactant.

____ 11. If #%.% g of sodium reacts with "2.% g of iron (III) o&ide, sodium acts as the limiting reactant.

____ 12. *ased on the mole ratios of the substances in a chemical reaction shown, determine the correct equation for

the chemical reaction+

ubstances -ole ratio

/* 3/2

/0 3/1*/0 2/1


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____ 13. 0alculate the mole ratio of all the substances represented in this figure+

____ 14. hich is true of the reaction shown below+

a. The mole ratio of this reaction is !//!.b. Two molecules of ubstance will be left o'er when this reaction goes to completion.

c. ubstance is the limiting reagent in this reaction.

d. The addition of more molecules of ubstance will not affect the amount of ubstance

that can be made.

____ 1. hich con'ersion factor will correctly complete this setup for finding the number of moles of 52required to

completely react with ".% grams of b+ The balanced chemical equation for the reaction is shown/

____ 1!. The table shows the mole ratios of potassium and bromine combined to form potassium bromide according to

the balanced reaction . In which trial(s) is the amount of bromine the limiting reagent+Trial 6umber -oles of 7 -oles of *r

1 1% 3

2 3 2

3 14 #

____ 1". 8old is reacted with chlorine gas according to the reaction 2 u 9 3 0l22 u0l3. :se the data in the table

to determine the percent yield of gold chloride.

-ass of 8old -ass of 0hlorine Theoretical ield ctual ield

3$.4 g 21.3 g + 3.2 g

____ 1#. ccording to this chemical reaction, calculate the number of grams of ;e produced from 14 moles of 54)29 7653+

____ 2%. =etermine the correct mole ratio for aluminum chloride to chlorine in the chemical reaction

l0l39 *r2l*r39 0l2+


3/18

____ 21. 0alculate how many moles of 7*r will be produced from " moles of *a*r2+ *a*r2 9 72547*r 9 *a54

____ 22. 0alculate how many moles of l would be produced from 2% moles of l253+ l253l 9 52

____ 23. =etermine how many moles of 0u are needed to react with .# moles of g653+

0u 9 2 g6530u(653)29 2 g

____ 24. hich is the number of moles of carbon dio&ide produced from the complete combustion of .42 moles of

ethanol+ 02

____ 2. hich is the correct number of moles of 65 that is produced from 13.2 moles of o&ygen gas in the presence

of e&cess ammonia+ 4 6


4/18

____ 32. hat is the mass of potassium chloride when !." g of potassium reacts with an e&cess of chlorine gas+ The

balanced chemical equation is/ .

____ 33. certain reaction has a "3.!? yield. If 3.# grams of the product were predicted by stoichiometry to be

made, what would the actual yield be+

____ 34. reaction was predicted to produce 32.4 grams of a compound. hen the product was measured, there were

only 2!.1 grams made. hat is the percent yield of this reaction+

____ 3.


5/18

____ 42.


6/18

1. ;ind the mole ratio of lead (II) nitrate to potassium nitrate in this chemical reaction.

7I 9 >b(653)2>bI29 7653

2. 2).

=etermine the reactant that is in e&cess.


7/18

!3. In a reaction, #2.%% g of sodium reacts with "4.%% g of ferric o&ide to form sodium o&ide and iron metal.

0alculate the mass of solid iron produced.

!4. In a reaction, 1%."! g of 0a053, 1%.1 g of articlesb. -olesc. -ass

"2. @epresent the reaction between Ainc and nitric acid in terms of/

a. >articles

b. -oles

c. -ass


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"3. In the equation for the combustion of butane, show that the law of conser'ation of mass is obser'ed. Interpret

the equation for the combustion of butane in terms of/

a. @epresentati'e particlesb. -oles

c. -ass

"4. 4.%%Eg sample of sil'er nitrate is mi&ed with .%% g of hydrochloric acid to form a white precipitate of

sil'er chloride. fter the solution is filtered and dried, a white precipitate of mass 33.% g is collected.

a. =etermine the limiting reactant.

b. =etermine the theoretical yield of sil'er chloride.c. =etermine the percent yield of sil'er chloride.


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Unit 8: Stoichiometry

Answer Section

TRUE/FALSE

1. 6/ T >T/ 1 =I;/ *loomFs De'el 1

6T/ *.3

2. 6/ ; >T/ 1 =I;/ *loomFs De'el 2

6T/ *.1 G *.2 G *.3

3. 6/ ; >T/ 1 =I;/ *loomFs De'el 3

6T/ *.1H G *.3

4. 6/ T >T/ 1 =I;/ *loomFs De'el 2

6T/ *.1 G *.2 G *.3

. 6/ ; >T/ 1 =I;/ *loomFs De'el 2

6T/ *.3

!. 6/ ; >T/ 1 =I;/ *loomFs De'el 2

6T/ *.3

". 6/ T >T/ 1 =I;/ *loomFs De'el 2

6T/ *.1 G *.2 G *.3

#. 6/ T >T/ 1 =I;/ *loomFs De'el 16T/ *.3

$. 6/ T >T/ 1 =I;/ *loomFs De'el 2

6T/ *.3

1%. 6/ ;

The percent yield is the ratio of the actual yield to the theoretical yield, and is e&pressed as a percent.

>T/ 1 =I;/ 1 @B;/ >age 3"%

5*/ 12.4.2 =etermine the percent yield for a chemical reaction.6T/ :0>.3 G *.3 T5>/ =etermine the percent yield for a chemical reaction.

7B/ Theoretical yield -0/ 1

65T/ The theoretical yield is the ma&imum amount of product that can be produced from a gi'en amount of

reactant.

11. 6/ ;

>T/ 1 =I;/ 2 @B;/ >age 3!"

5*/ 12.3.1 Identify the limiting reactant in a chemical equation.

6T/ *.3 T5>/ Identify the limiting reactant in a chemical equation.

7B/ Dimiting reactant -0/ 365T/ The actual ratio (%.12$) is less than the required ratio (%.1!!). Thus, iron (III) o&ide is the limiting

reactant.

MULTPLE !"#!E

12. 6/ = >T/ 1 =I;/ *loomFs De'el 26T/ :0>.2 G *.3


10/18

13. 6/ * >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.2 G *.1 G *.2

14. 6/ * >T/ 1 =I;/ *loomFs De'el 4

6T/ :0>.2 G *.3

1. 6/ 0 >T/ 1 =I;/ *loomFs De'el 46T/ :0>.2 G *.3

1!. 6/ = >T/ 1 =I;/ *loomFs De'el 4

6T/ :0>.2 G *.3

1". 6/ 0 >T/ 1 =I;/ *loomFs De'el 4

6T/ :0>.2 G *.3

1#. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ *.3

1$. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ *.1 G *.3

2%. 6/ * >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

21. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.322. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3

23. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3

24. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

2. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

2!. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

2". 6/ = >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

2#. 6/ = >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

2$. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

3%. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

31. 6/ * >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

32. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

33. 6/ = >T/ 1 =I;/ *loomFs De'el 36T/ :0>.2 G :0>.3 G *.3

34. 6/ >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

3. 6/ * >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

3!. 6/ 0 >T/ 1 =I;/ *loomFs De'el 3

6T/ :0>.3 G *.3

3". 6/ 0


11/18

The equation for the combustion of propane is .

Feedback

A =i'ide the unCnown moles of carbon dio&ide by the Cnown moles of propane.B -ultiply the Cnown number of moles of propane by the mole ratio.C 0orrectJD *alance the equation correctly.

>T/ 1 =I;/ 2 @B;/ >age 3$

5*/ 12.2.2 :se the steps to sol'e stoichiometric problems. 6T/ :0>.1 G :0>.3 G *.3

T5>/ :se the steps to sol'e stoichiometric problems.

7B/ toichiometric moleEtoEmole con'ersion -0/ 3

3#. 6/ =

The balanced chemical equation is .

Feedback

A -ultiply the number of moles of water by the molar mass of water.B

-ultiply the number of moles by the mole ratio.C *alance the equation correctly.D 0orrectJ

>T/ 1 =I;/ 3 @B;/ >age 3!%5*/ 12.2.2 :se the steps to sol'e stoichiometric problems. 6T/ :0>.1 G :0>.3 G *.3


7B/ toichiometric moleEtoEmass con'ersion -0/ 3

3$. 6/

The balanced chemical equation is .

Feedback

A 0orrectJB 0alculate the mass of 70l using the molar mass as a con'ersion factor.C *alance the equation correctly.D 0on'ert the grams of 70l to moles using the in'erse of molar mass as the con'ersion

factor.

>T/ 1 =I;/ 3 @B;/ >age 3!1



7B/ toichiometric massEtoEmass con'ersion -0/ 3

4%. 6/ 0

The mole ratio of carbon dio&ide to ethanol is 2/1.

Feedback

A =i'ide the unCnown moles by the Cnown moles.B -ultiply the Cnown number of moles by the mole ratio.C 0orrectJD *alance the equation correctly.

>T/ 1 =I;/ 1 @B;/ >age 3$


12/18



7B/ toichiometric moleEtoEmole con'ersion -0/ 3

41. 6/ =

The molar mass of citric acid is 1$2.%44 g.

Feedback

A =i'ide the unCnown moles by the Cnown moles.B -ultiply the Cnown number of moles by the mole ratio and the molar mass.C *alance the equation correctly.D 0orrectJ

>T/ 1 =I;/ 2 @B;/ >age 3!%



7B/ toichiometric moleEtoEmass con'ersion -0/ 3

42. 6/ =

>ercent yield (actual yieldKtheoretical yield)

1%%

Feedback

A -ultiply the yield by 1%% to calculate the percent yield.B =i'ide the actual yield by the theoretical yield.C The molar mass is incorrect.D 0orrectJ

>T/ 1 =I;/ 3 @B;/ >age 3"1

5*/ 12.4.2 =etermine the percent yield for a chemical reaction.

6T/ :0>.3 G *.3 T5>/ =etermine the percent yield for a chemical reaction.

7B/ >ercent yield -0/ 3

43. 6/ =>ercent yield (actual yieldKtheoretical yield) 1%%

Feedback

A -ultiply the number of moles by the molar mass to obtain the theoretical yield.B The mole ratio is incorrect.C The molar mass is incorrect.D 0orrectJ

>T/ 1 =I;/ 3 @B;/ >age 3"1




!#MPLET#N

44. 6/ chlorine

>T/ 1 =I;/ 2 @B;/ >age 3!

5*/ 12.3.1 Identify the limiting reactant in a chemical equation.


13/18

6T/ *.3 T5>/ Identify the limiting reactant in a chemical equation.

7B/ Dimiting reactant -0/ 3

4. 6/ e&cess

>T/ 1 =I;/ 2 @B;/ >age 3!

5*/ 12.3.2 Identify the e&cess reactant and calculate the amount remaining after the reaction is complete.

6T/ :0>.3 G *.3T5>/ Identify the e&cess reactant and calculate the amount remaining after the reaction is complete.

7B/ B&cess reactant -0/ 1

S"#RT ANS$ER

4!. 6/

2 9 3* 0 9 !=

>T/ 1 =I;/ *loomFs De'el 3 6T/ :0>.2 G *.1 G *.3

4". 6/

) molar mass of ubstance *) mole ratio between substance and substance

0) molar mass of substance

>T/ 1 =I;/ *loomFs De'el 2 6T/ :0>.2 G :0>.3 G *.3

4#. 6/

i& molecules of diatomic ubstance , 4 molecules of ubstance .

>T/ 1 =I;/ *loomFs De'el 6T/ :0>.2 G *.3

4$. 6/

The mole ratio is re'ersedL it should be 4 moles of ;eK3 moles 52. This is because the number of moles of

o&ygen must be on the bottom of the fraction in order to cancel out the moles of o&ygen in the pre'ious

fraction (molar mass of o&ygen).

>T/ 1 =I;/ *loomFs De'el ! 6T/ :0>.2 G :0>.3 G *.3

%. 6/n e&cess reagent will accomplish two things. It will ensure that the reaction goes to completion, since all of

the limiting reagent will be able to be used. It will also help to maCe the reaction go faster.

>T/ 1 =I;/ *loomFs De'el 2 6T/ :0>.2 G *.3

1. 6/

1/2


14/18

>T/ 1 =I;/ *loomFs De'el 3 6T/ *.3

2. 6/

3%.! molesL T/ 1 =I;/ *loomFs De'el 3 6T/ :0>.3 G *.3

3. 6/

a) -agnesium is limiting and nitrogen is e&cessL they react in a 3/1 molar ratio so there is proportionately less

magnesium.b) 2 moles of product are formedL the mole ratio for the entire reaction is 3/1/1

c) ince 2 moles of nitrogen must be used, there will be 4 moles left o'er.

>T/ 1 =I;/ *loomFs De'el 4 6T/ :0>.2 G *.3

4. 6/

3%.3 grams

The bromine is the limiting reagentL %.42# moles of chlorine will be produced.

>T/ 1 =I;/ *loomFs De'el 3 6T/ :0>.3 G *.3

. 6/

The theoretical yield would be 1%!." grams. Therefore, the percent yield is ("".1K1%!.")&1%%, or "2.3?.


!. 6/

ith a percent yield of #3.1? and an actual yield of 43. grams, the theoretical yield must be at least (43.

g)K(%.#31) M 2.3 grams of chlorine gas. :sing this 'alue in stoichiometry, we find that 2.3 grams of chlorinerequires !!.4 grams of aluminum chloride.


". 6/

toichiometry is a study of quantitati'e relationships between the amounts of reactants used and the products

formed by a chemical reaction.

>T/ 1 =I;/ 1 @B;/ >age 34

5*/ 12.1.1 Identify the quantitati'e relationships in a balanced chemical equation.

6T/ :0>.3 G *.3 T5>/ Identify the quantitati'e relationships in a balanced chemical equation.

7B/ toichiometry -0/ 1

#. 6/

mole ratio is the ratio between the numbers of moles of any two substances in a balanced chemical

equation.

>T/ 1 =I;/ 1 @B;/ >age 3!

5*/ 12.1.2 =etermine the mole ratios from a balanced chemical equation.

6T/ :0>.3 G *.3 T5>/ =etermine the mole ratios from a balanced chemical equation.7B/ -ole ratio -0/ 1

$. 6/The balanced chemical equation is/

The possible mole ratios are/


15/18

>T/ 1 =I;/ 2 @B;/ >age 3!

5*/ 12.1.2 =etermine the mole ratios from a balanced chemical equation.6T/ :0>.3 G *.3 T5>/ =etermine the mole ratios from a balanced chemical equation.

7B/ -ole ratio -0/ 3

!%. 6/

The four steps to sol'e stoichiometric problems are/

a. rite a balanced chemical equation.

b. =etermine the moles of a gi'en substance using a massEtoEmole con'ersion.

c. =etermine the moles of an unCnown substance from the moles of the gi'en substance.

d. =etermine the mass of an unCnown substance from the moles of the unCnown substance using a moleEtoEmass con'ersion.

>T/ 1 =I;/ 1 @B;/ >age 3!3

5*/ 12.2.1 B&plain the sequence of steps used in sol'ing stoichiometric problems.

6T/ :0>.1 G :0>.3 G *.3

T5>/ B&plain the sequence of steps used in sol'ing stoichiometric problems.7B/ :sing stoichiometry -0/ 2

PR#%LEM

!1. 6/##.!1?

>T/ 1 =I;/ 3 @B;/ >age 3"1




65T/ >ercent yield M (actual yieldKtheoretical yield) N 1%%

!2. 6/>hosphorus is the reactant that is in e&cess.

>T/ 1 =I;/ 3 @B;/ >age 3!"

5*/ 12.3.2 Identify the e&cess reactant and calculate the amount remaining after the reaction is complete.6T/ :0>.3 G *.3

T5>/ Identify the e&cess reactant and calculate the amount remaining after the reaction is complete.

7B/ B&cess reactant -0/ 365T/ The actual ratio is less than the required ratio. The number of moles of chlorine needed in the reaction

is 1%, but only 1.2$ moles of chlorine gas are a'ailable. Thus, chlorine acts as the limiting reactant and

phosphorus as the e&cess reactant.

!3. 6/

1.3! g


16/18

>T/ 1 =I;/ 3 @B;/ >age 3!"

5*/ 12.3.3 0alculate the mass of a product when the amounts of more than one reactant are gi'en.

6T/ :0>.3 G *.3

T5>/ 0alculate the mass of a product when the amounts of more than one reactant are gi'en.

7B/ >roduct mass calculation -0/ 3

65T/ ;irst, calculate the actual ratio. Then, con'ert the number of moles of ferric o&ide to the number ofmoles of iron.

!4. 6/

23.! g

>T/ 1 =I;/ 3 @B;/ >age 3"1

5*/ 12.4.1 0alculate the theoretical yield of a chemical reaction from data.

6T/ :0>.3 G *.3 T5>/ 0alculate the theoretical yield of a chemical reaction from data.

7B/ Theoretical yield -0/ 3

65T/ The theoretical yield is calculated by multiplying the number of moles of calcium chloride

he&ahydrate by the molar mass.

!. 6/

1#?

>T/ 1 =I;/ 3 @B;/ >age 3"1




65T/ >ercent yield M (actual yieldKtheoretical yield) N 1%%

ESSA&

!!. 6/

The percent yield is ""."?. There may be procedural reasons why the yield is not 1%%?, such as a precipitatebeing left on filter paper or otherwise left behind. Diquids might sticC to their containers or e'aporate. ;inally,

there may be other reactions occurring at the same time, remo'ing some of the reactants and pre'enting them

from forming the desired product.

>T/ 1 =I;/ *loomFs De'el 4 6T/ :0>.3 G *.3

!". 6/

;e2539 302;e 9 305

>articles/ 2 atoms ;e, 3 atoms 5, 3 atoms 0 as reactantsL 2 atoms ;e, 3 atoms 5, 3 atoms 0 as products

-oles/ 2 moles ;e, 3 moles 5, 3 moles 0 as reactantsL 2 moles ;e, 3 moles 5, 3 moles 0as products

-ass/ 1(1$.!$1) 9 3(12.%1) M 2 (.#4") 9 3 (#4.%2")L 1$."21 g M 1$."21

>T/ 1 =I;/ *loomFs De'el ! 6T/ :0>.2 G *.3

!#. 6/

a) 4l(s) 9 352(g)2l253(s)

b) The limiting reagent is the aluminum.c) 1%.3 grams of aluminum o&ide can be produced.


17/18


!$. 6/

The molar amounts in this reaction are $.1" moles nitrogen and 41. moles hydrogen, based on the molar

masses of each compound. The mole ratio for the entire reaction is 1/3/2. ince 41. moles is more than 3

times larger than $.1" moles, the nitrogen is the limiting reagent. It will only react with 2". moles of

hydrogen, lea'ing 14 moles of hydrogen in e&cess.

>T/ 1 =I;/ *loomFs De'el 4 6T/ :0>.2 G *.3

"%. 6/

a) 2 6a5< 9 -g(653)22 6a6539 -g(5T/ 1 =I;/ *loomFs De'el 6T/ :0>.2 G *.3

"1. 6/

a. 1 molecule 629 3 molecules


18/18

>T/ 1 =I;/ 3 @B;/ >age 3"1




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