Chapter 11Stoichiometry

Stoichiometry

• The study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction

• Law of Conservation of Mass

How Much?

• If you were making cookies, you would need the right amount of each ingredient to get the two dozen cookies the recipe makes.

• When performing a chemical reaction it is important to know how much of each reactant is needed to get the desired amount of product(s).

Balanced Equations

• The first step in determining anything in stoichiometry is balancing the chemical equation – Let’s Review!

N2 + H2 → NH33 2

Practice

1. CH4 + O2 → CO2 + H2O

2. NaOH + CaBr2 → Ca(OH)2 + NaBr

3. H2 + Cl2 → HCl

4. Li + H2O → LiOH + H2

Types of Relationships

• Particle to Particle• Mole to Mole• Mass to Mass

C3H8(g) + 5O2(g)→ 3CO2(g)+ 4H2O(g)

One molecule of propane gas reacts with five molecules of oxygen gas to form three molecules of carbon dioxide gas and four molecules of water vapor.

One mole of propane gas reacts with five moles of oxygen gas to form three moles of carbon dioxide gas and four moles of water vapor.

44.10 g of propane gas reacts with 159.99 g of oxygen gas to form 132.03 g of carbon dioxide gas and 72.06 g of water vapor.

Practice

5. 2H2O(g) → 2H2(g)+ O2(g)

6. 2NaCl(aq) + MgBr2(aq) → MgCl2(s)+ 2NaBr(aq)

7. CaO(s) + CO2(g) → CaCO3(s)

Mole Ratio

• The ratio between the numbers of moles of any two substances in a balanced chemical equation. Taken from the coefficients of the balanced equation.

2K(s) + Br2(l) → 2KBr(s)

2 mol K 2 mol K1 mol Br2 2 mol KBr2 mol KBr 2 mol KBr1 mol Br2 2 mol Br21 mol Br2 1 mol Br2

2 mol K 2 mol KBr

Mole Ratio

• For reactions with more reactants and products there are more mole ratios

BaS(aq) + 2KOH(aq) → Ba(OH)2 + K2S

1 mol BaS 1 mol BaS 1 mol BaS2 mol KOH 1 mol Ba(OH)2 1 mol K2S

2 mol KOH 2 mol KOH 2 mol KOH1 mol BaS 1 mol Ba(OH)2 1 mol K2S

Practice

8. H2 + Cl2 → HCl

9. H2O → H2 + O2

10. C2H10 + O2 → CO2 + H2O

11. LiOH + CaBR → Ca(OH)2 + LiBr

Mole to Mole Conversions

N2 + 3H2 → 2NH3

• If we have one mole of nitrogen gas, we need three moles of hydrogen gas in order to produce two moles of ammonia.

• What if we have 0.012 mol of nitrogen gas?

Mole to Mole Conversions

N2 + 3H2 → 2NH3

• What if we have 0.012 mol of nitrogen gas?• Use mole ratios

3 mol H2

1 mol N2

2 mol NH3

1 mol N2

0.012 mol N2 × = 0.036 mol H2

0.012 mol N2 × = 0.024 mol H2

Practice

12. 0.6 mol hydrogen gas

H2 + Cl2 → HCl

13. 1.45 mol oxygen gas

C2H10 + O2 → CO2 + H2O

14. 1.32 mol hydrogen gas

H2O → H2 + O2

15. 0.24 mol lithium hydroxide

LiOH + CaBr2 → Ca(OH)2 + LiBr

N2(g) + 3H2(g) → 2NH3(g)• What if we have 5.20 mol of nitrogen gas? How

many grams of ammonia are produced in an excess of hydrogen gas?

• Use mole ratios, then molar mass2 mol NH3

1 mol N2

Mole to Mass Conversions

5.20 mol N2 = 15.6 mol NH3×

15.6 mol NH3 17.03 g1 mol NH3

× = 266 g NH3

4Al(s) + 3O2(g) → 2Al2O3(s)• What if we have 0.42 mol of oxygen gas? How many

grams of aluminum oxide are produced in an excess of aluminum solid?

• Use mole ratios, then molar mass2 mol Al2O3

3 mol O2

Mole to Mass Conversions

0.42 mol O2 = 0.28 mol Al2O3×

0.28 mol Al2O3 101.96 g1 mol Al2O3

× = 29 g Al2O3

Practice

16. 0.6 mol chlorine gas

Na(s) + Cl2(g) → NaCl(s)

17. 12 mol oxygen gas

P(s) + O2(g) → P2O5(s)

18. 3.05 mol oxygen gas

CH4(g) + O2(g) → CO2(g) + H2O(g)

19. 0.96 mol sodium bromide

NaBr(s) + CaF2(s) → CaBr2(s) + NaF(s)

N2(g) + 3H2(g) → 2NH3(g)• What if we have 27 g of nitrogen gas? How many grams of

ammonia are produced in an excess of hydrogen gas?• Use molar mass, then mole ratios, then molar mass

2 mol NH3

1 mol N2

Mass to Mass Conversions

0.96 mol N2 = 1.92 mol NH3×

1.92 mol NH3 17.03 g1 mol NH3

× = 33 g NH3

1 mol N2

28.01 g N2

27 g N2 = 0.96 mol N2×

4Al(s) + 3O2(g) → 2Al2O3(s)• What if we have 14.3 g of oxygen gas? How many grams of

aluminum oxide are produced in an excess of aluminum solid?• Use molar mass, mole ratios, then molar mass

2 mol Al2O3

3 mol O2

Mass to Mass Conversions

0.447 mol O2 = 0.298 mol Al2O3×

0.298 mol Al2O3 101.96 g1 mol Al2O3

× = 30.4 g Al2O3

1 mol O2

32.00 g O2

14.3 g O2 = 0.447 mol O2×

Practice

20. 46 g chlorine gas

Na(s) + Cl2(g) → NaCl(s)

21. 31.7 g oxygen gas

P(s) + O2(g) → P2O5(s)

22. 3.05 g oxygen gas

CH4(g) + O2(g) → CO2(g) + H2O(g)

23. 6.41 g sodium bromide

NaBr(s) + CaF2(s) → CaBr2(s) + NaF(s)