Chapter 11: Stoichiometry

Chapter 11: Stoichiometry

Honors Chemistry

Stoichiometry - the study of the quantitative relationship between the amounts of reactants used and the amounts of products formed in a chemical reaction.

11.1 Defining Stoichiometry

11.1 Defining Stoichiometry

Use coefficients from the balanced chemical equation as your conversion factor

2K (s) + 2H

O (l) →2KOH (aq) + H

(g) 2 mol K = 2 mol H

O = 2 mol KOH = 1 mol H

1. How many moles of H

can be produced from 5 moles of K?

2. How many moles of H

O are needed to produce 84.6 mole of H

?

11.2 Stoichiometric Calculations

Use coefficients from the balanced chemical equation as your conversion factor

Convert to moles first, then use (mole/mole), then convert to grams

NH

NO

→N

O+ 2H

O

3. How many grams of H

O are produced from 14.87 g of NH

NO

4. How many grams of NH

NO

are needed to produce 249.6 g of N

O?

Mass to mass problems

Pb(NO ₃ ) ₂ + 2KI → PbI ₂ + 2KNO ₃

5. How many grams of PbI ₂ will be produced from 149.0 g of KI?

6. How many grams of KNO ₃ will be produced if 54.6 grams of PbI ₂ are also produced?

Mass to mass problems

Density will be listed as g/mL or g/L.

Use the conversion _______g= 1mL or ______g = 1L EXAMPLE: the density of water is .99705 g/mL so use

.99705 g = 1 mL in your ()

2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

7. How many mL of water are produced when 74.90 g of Mg(OH)₂ are reacted?

8. How many grams of Mg₃(PO₄)₂ are produced when 105.2 mL of water are also produced?

Problems using Density (not in book)

2C

H

+ 15O

→ 12CO

+ 6H

O Density of water is .99705 g/mL

9. How many L of O

gas at STP are needed to produce 46.99 g of water?

10. If 100.0 g of C

H

are reacted, how many mL of water are produced?

Putting it all together

2C

H

+ 15O

→ 12CO

+ 6H

O Density of water is .99705 g/mL

11. 4.6 x 10

molecules of CO

are produced, how many L of O

gas at STP were reacted?

12. How many molecules of C

H

were reacted in order to produce 64.9 L of CO

at STP?

Putting it all together

Problems involving heat (not in book)

Heat can be thought of as just another reactant (endothermic) or product (exothermic) in a chemical equation and written into the conversion factors

with the coefficients

N₂(g)+ 3H₂(g) →2NH₃(g)+ 92.4 kJ 1 mol N₂ = 3 mol H₂ = 2 mol NH₃ = 92.4 kJ

13. How many grams of N₂ are needed to produce 150.9 kJ of heat?

14. How many kJ of heat are produced when 59.7 L of H₂ gas react at STP?

15. How much energy is released when 6.49 x 10²² molecules of NH₃ are produced?

11.3 Limiting Reactants

Sometime you have known quantities for two reactants, so which one do you use at the beginning of a stoichiometry

problem?

TWO WAYS

1. Find which one is the limiting reactant and solve stoichiometry problem starting with the limiting reactant

2. Do the stoichiometry problem two times; starting with each reactant, the one with the SMALLEST product is the

correct one and the limiting reactant.

16Ag + S

→8Ag

S

16. When 4.00 grams of silver react with 4.00 grams of sulfur, what mass of silver sulfide is produced?

STEP 1 – convert both reactants to moles

STEP 2 – Answer the following question – “if I used up all the moles of reactant 1, how many moles of reactant 2

would I need?” You do this using a mole to mole

stoichiometry problem. The answer to this is the number of moles of reactant 2 needed.

11.3 Limiting Reactants

STEP 3 – Compare moles of reactant 2 that you have (from step 1) to the number of moles of reactant 2 that is needed (from step 2). If the number of moles of reactant 2 that you have is greater than the number of moles of reactant 2 needed, than reactant 2 is in excess, reactant 1 is the limiting reactant.

If the number of moles of reactant 2 that you have is less than the number of moles of reactant 2 needed, than reactant 2 is the limiting reactant.

SUMMARY – look at reactant 2

If NEEDED is less than HAVE - reactant 1 is the limiting reactant.

If NEEDED is bigger than HAVE – reactant 2 is the limiting reactant

For this problem __________________________ is the limiting reactant

STEP 4 – Continue with stoichiometry problem to find the answer to the problem using the limiting reactant as a starting point.

11.3 Limiting Reactants

17. The reaction between solid sodium and iron (III) oxide is one in a series of reaction that inflates and automobile airbag

6Na (s) + Fe

O

(s) →3Na

O(s) + 2Fe(s)

What mass of solid iron is produced when 50.0 grams of sodium is reacted with 50.0 grams of iron (III) oxide?

18. What mass of Na

O is produced when 25 grams of sodium is reacted 50.0 grams of Fe

O

?

11.3 Limiting Reactants

4Al (s) + 3O

(g) →Al

O

(s)

19. What mass of Al

O

is formed from 100. grams of Al and 150. grams of O

?

20. What mass of Al

O

is formed from 35.0 grams of Al and 35.0 grams of O

?

11.3 Limiting Reactants

11.4 Percent Yield

Main Idea: Percent yield is a measure of the efficiency of a chemical reaction

Stoichiometry tells us what should happen, but in the lab often times we don’t make as much product as we

calculated.

% yield = (actual yield/theoretical yield)*100%

11.4 Percent Yield

Carbon tetrachloride, CCl₄, is a solvent that was once used in large quantities in dry cleaning. Because it is a dense liquid that does not burn, it was also used in fire extinguishers. Unfortunately, its use was discontinued because it was found to be a carcinogen. It was manufactured by the following reaction:

CS₂ + 3Cl₂ →CCl₄ + S₂Cl₂

The reaction was economical because the byproduct disulfur dichloride, S₂Cl_2, could be used by industry in the manufacture of rubber products and other materials.

21. What is the percent yield of CCl₄ if 719 g is produced from the reaction of 410. g of CS₂.

11.4 Percent Yield

22. If 67.5 g of Cl

are used in the reaction and 39.5 g of S

Cl

is produced, what is the percent yield?

23. If the percent yield of the industrial process is 83.3%, how many grams of CS

should be reacted to obtain 5.00 x 10

g of CCl

? How many grams of S

Cl

will be produced, assum ing the same yield for that product?

11.4 Percent Yield

Nitrogen dioxide, NO₂, can be converted to dinitrogen pentoxide, N₂O₅, by reacting it with ozone, O₃, The reaction of NO₂ takes place according to the following equation:

2NO₂ (g) + O3 (g) →N₂O₅ (s or g) + O₂ (g)

24. Calculate the percent yield for a reaction in which 0.38 g of NO₂ reacts and 0.36 g of N₂O₅ is recovered.

25. What mass of N₂O₅ will result from the reaction of 6.00 mol of NO₂ if there is a 61.1 % yield in the reaction?

11.4 Percent Yield

26. In the past, hydrogen chloride, HCl, was made using the salt-cake method as shown in the following equation:

2NaCl (s) + H

SO

(aq) →Na

SO

(s) + 2HCl (g)

If 30.0 g of NaCl and 0.250 mol of H

SO

are available, and 14.6 g of HCl is made, what is the percent yield?

11.4 Percent Yield