stoichiometry

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Stoichiometry

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Stoichiometry. Stoichiometry Steps. Balance the equation;. Is it balanced?. Is it balanced?. Mole Box. Grams A. Grams B. Molar Mass of A. Molar Mass of B. Mol A. Mol B. Mole Ratio (Balanced Equation). 2. Grams to Moles (Divide by Molar Mass) *Don’t forget to use sig figs!*. - PowerPoint PPT Presentation

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Page 1: Stoichiometry

Stoichiometry

Page 2: Stoichiometry
Page 3: Stoichiometry
Page 4: Stoichiometry

Stoichiometry Steps

1. Balance the equation;Is it

balanced?

Is it balanced?

Page 5: Stoichiometry

Mole Box

Grams A

Mol A Mol B

Grams B

Molar Mass of A

Mole Ratio(Balanced Equation)

Molar Mass of B

Page 6: Stoichiometry

2. Grams to Moles (Divide by Molar Mass)*Don’t forget to use sig figs!*

27.1 grams of O2 X 1 mole O2 =

32 grams O2

0.847 moles of O2

A sample contains 27.1 g of oxygen. How many moles of oxygen are in this sample?

This is the molar mass of Oxygen. (16.00g X 2)= 32.00g of O2 *THIS IS A ONE STEP PROBLEM!

*

Page 7: Stoichiometry

3. Moles to Moles ( Multiply by Mole Ratio)

If you started with 3.50 mol of hydrogen, how many moles of oxygen could be formed?

3.50 mol H2 X 1 mole O2 = 1.75 moles of O2 2 moles H2

Page 8: Stoichiometry

4. Mole to Grams (Multiply by Molar Mass)

If you needed to produce 12.2 g of water, how many grams of hydrogen would you need?

12.2 moles of H2O X 2 moles H2 X 2 grams H2 = 24.4 grams of H2 2 moles H2O 1 mole H2

*THIS IS A TWO STEP PROBLEM!*

Page 9: Stoichiometry

Molarity

The concentration of a solution in

mol/L = moles of solute 1 L of solution

Page 10: Stoichiometry

Finding Moles of a Solution

Use the formula;

cv = nConcentration(mol/L) X Volume(L) = Moles(mol)

Page 11: Stoichiometry

Mole Box 2Grams A

Mol A Mol B

Grams B

Litres A Litres B

Molar Mass of A

Mole Ratio(Balanced Equation)

Molar Mass of B

Concentration of A (mol/L)

Concentration of B (mol/L)

Page 12: Stoichiometry

Example(Litres A to Grams B)

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

15.0 mL of a 1.5 mol/L NaOH solution gives how many grams of H2O?

0.0150 L NaOH X 1.5 mol NaOH X 2 mol H2O X 18g H2O = 1 L NaOH 2 mol NaOH 1 mol H2O

CONVERT! 1 L = X L X= 0.0150 L 1000 mL 15.0 mL 0.4

05 g H2O

*THIS IS A THREE STEP PROBLEM!*

Page 13: Stoichiometry

Stoichiometry with Gases

The stoichiometry of gases is based on the ratio between the quantities of gas involved in a chemical reaction. This method is used to predict the quantity

of a reactant or product involved in a chemical reaction in which one of the compounds is

a gas.

Page 14: Stoichiometry

Example 1(Litres A to Litres B)

If 7.00 L of H2 gas reacts with Cl2 gas, how many litres of HCl gas would form?

H2 + Cl2 2HCl

7.00 L H2 X 2 L HCl = 14.0 L HCl

1 L H2

Page 15: Stoichiometry

Example 2(Grams A to Litres B)

Mg(s) + HCl(aq) MgCl2(aq) + H2(g)

If 17.00 g of Mg reacts, how many millilitres of H2 will form?

17.00 g Mg X 1 mol Mg X 1 mol H2 X 22.4 L H2 = 15.70 L of H2 24.31 g Mg 1 mol Mg 1 mol H2

Convert this answer to millilitres!

1 L = 15.70 L X= 15 700 mL of H21000 mL X mL

Page 16: Stoichiometry

Classifying Reactions

1. Combustion= Burning with OxygenTips:If C is in the reactants, CO2 will be a product.

Example:• C + O2 CO2If H2 is in the reactants, H2O will be a product.

Example:• 2H2 + O2 2H2O• CH4 + 2O2 CO2 + 2H2O

Page 17: Stoichiometry

2. Synthesis

Simpler reactants combine to make more complex products.

Examples:• Mg + Cl2 MgCl2• CO2 + H2O H2CO3

Page 18: Stoichiometry

3. Decomposition

Decomposition is when complex Reactants break down to become

simpler molecules.

Examples: • H2CO3 H2O + CO2• BF3NH3 BF3 + NH3

Page 19: Stoichiometry

4. Single Displacement

An element and a compound are required for single displacement. The

element in the reactants will “replace” or “displace” one of the

elements in the compound. *It is important to remember that

metals replace metals only, and non-metals replace non-metals only.

Examples:• A + B B + AC• Na(s) + H2O(l) NaOH(aq) + H2(g)

Page 20: Stoichiometry

Example of a Reaction

Burning of methane releases 275 kJ of energy per mole of methane being burned. How many grams of methane must burn in order to produce 1500 kJ of energy?

CH4 + 2O2 CO2 + 2H2O1500 kJ X 1 mole CH4 X 16.04 g CH4 275 kJ 1 mole CH4

= 88

g of

CH4

C = 12.01 12.01 + 1.01(4) = 16.04 g of CH4 H = 1.01

What type of reaction has occurred? Combustion reaction.

Page 21: Stoichiometry

Bibliography

• Couture, Ivan, Marie-Eve Lacombe-Harvey, and Genevieve Levasseur-Thériault. Quantum Chemistry Student Textbook. Ed. Marie-Eve Robitaille, Isabel Rusin, and Colleen Ovenden. Trans. Jacquie Charlton, Cristina Cusano, Natasha DeCruz, Joann Egar, and Gwen Schulman. Montreal: Cheneliere Education, 2011. Print.

• Doty, Wendy L. "Stoichiometry Notes." Downey Unified School District. N.p., 2006. Web. 10 Oct. 2012. <http://www.dusd.net/staff/rcramm/Chemistry/Unit_8/Stoichiometery%20Notes.pdf>.