1. Chapter 9 Stoichiometry
9.2
chemical calculations

2. Things you will learn
You will be able to construct mole ratios from balanced chemical equations and be able to apply these in mole-mole calculations.
You will be able to calculate stoichiometric quantities (mass, volume, representative particles, and moles) from balanced equations.

3. Baking soda experiment
Baking soda decomposes with heat to form sodium carbonate, carbon dioxide, and water
The skeleton equation is:
NaHCO Na2CO3 + CO2 + H2O

4. Baking soda experiment
Baking soda decomposes with heat to form sodium carbonate, carbon dioxide, and water
The balanced equation is:
2NaHCO Na2CO3 + CO2 + H2O
2 Na 2
2 C 2
6 O 6
2 H 2

5. What we want to find out is how much baking soda will we have to heat in order to fill a liter Coke bottle half-full.
Since we have a balanced equation, we can use the molar road map to find out just how much we should use.
We will start by putting the given quantity on the left and the units of what we want to find on the right, and filling in the middle with conversion factors.

6. Mole-mole ratio from balanced equation
Given quantity
Wanted quantity

7. Given quantity
Ratio from balanced equation
.5 L CO2
1 mol CO2
22.4 L CO2
2 mol NaHCO3
1 mol CO2
84 g NaHCO3
1 mol NaHCO3
_ g NaHCO3
Wanted quantity
3.75 g NaHCO3

8. So if we put 3.75 grams of NaHCO3 in a test tube and heated it, we should get ½ liter of CO2 given off.
What else is given off that might mess up our calculations?
2NaHCO Na2CO3 + CO2 + H2O

9. What else is given off that might mess up our calculations?
So if we put 3.75 grams of NaHCO3 in a test tube and heated it, we should generate ½ liter of CO2.
What else is given off that might mess up our calculations?
2NaHCO Na2CO3 + CO2 + H2O
These are solids
These are BOTH gases!

10. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
Using this balanced equation, what are the possible mole ratios we can use as conversion factors?
Remember, conversion factors are always equal to 1 and so may be used upside down, with the numerator in the denominator, and vice versa

11. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
Using this balanced equation, what are the possible mole ratios we can use as conversion factors?
4 moles Al
3 moles O2
3 moles O2
2 moles Al2O3
4 moles Al
2 moles Al2O3
3 moles O2
4 moles Al
2 moles Al2O3
3 moles O2
2 moles Al2O3
4 moles Al

12. We use mole-mole conversions (ratios from balanced equations) to solve problems like:
How many moles of oxygen are required to react completely with 14.8 mole of aluminum?
We always work from our known (given) quantity towards our unknown (wanted) quantity.
14.8 mol Al
3 mol O2
4 mol Al
_ mol O2
mole-mole ratio
given
unknown

13. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
How many moles of Al2O3(s) would we get if we only used 1 mole of Al(s)?
1 mol Al
2 mol Al2O3
4 mol Al
_ mol Al2O3
mole-mole ratio
given
unknown

14. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
How many moles of Al2O3(s) would we get if we only used 1 mole of O2(g)?
1 mol O2
2 mol Al2O3
3 mol O2
_ mol Al2O3
mole ratio
given
unknown

15. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
How many moles of O2(g) would it take to make 5 moles of Al2O3(s) ?
5 mol Al2O3
3 mol O2
2 mol Al2O3
_ mol O2
mole ratio
given
unknown

16. Mole-mole calculations
4Al(s) + 3O2(g) Al2O3(s)
How many moles of Al(s) would it take to make 5 moles of Al2O3(s) ?
14.8 mol Al
3 mol O2
4 mol Al
_ mol O2
mole ratio
given
unknown

17. Mass-mass calculations
The number of moles of a reactant or product can be gotten directly from the coefficients of a balanced equation.
The amount in grams must be gotten from the molar mass of each. This adds a couple more conversion factors, but they’re easy to deal with.

18. Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of hydrogen with an excess of nitrogen
N2(g) + H2(g) NH3(g) (skeleton equation)

19. Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of hydrogen with an excess of nitrogen
N2(g) + 3H2(g) NH3(g) (balanced equation)

20. Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of hydrogen with an excess of nitrogen
N2(g) + 3H2(g) NH3(g)
5.4 g H2
given
Gram-mole conversion
Mole-mole conversion
Gram-mole conversion
_ g NH3
unknown

21. Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of hydrogen with an excess of nitrogen
N2(g) + 3H2(g) NH3(g)
5.4 g H2
1 mol H2
2 g H2
2 mol NH3
3 mol H2
17 g NH3
1 mol NH3
given
Gram-mole conversion
Mole-mole conversion
Gram-mole conversion
_ g NH3
unknown

22. Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of hydrogen with an excess of nitrogen
N2(g) + 3H2(g) NH3(g)
5.4 g H2
1 mol H2
2 g H2
2 mol NH3
3 mol H2
17 g NH3
1 mol NH3
given
Gram-mole conversion
Mole-mole conversion
Gram-mole conversion
30.6 g NH3
unknown

23. Mole-mole conversions Ammonia burns in oxygen forming nitrogen oxide and water. The skeleton formula is: NH3(g) + O2(g) NO(g) + H2O(g) Balance this equation and determine how many moles of ammonia are consumed if 1.35 moles of oxygen are used.

24. 4 N 4 12 H 12 10 O 10 1.35 mol O2 4 mol NH3 5 mol O2 _mol NH3
4NH3(g) + 5O2(g) NO(g) + 6H2O(g)
4 N 4
12 H 12
10 O 10
1.35 mol O2
4 mol NH3
5 mol O2
_mol NH3

25. Mass-mole conversions Phosphorus reacts with chlorine to form phosphorus pentachloride. The skeleton formula is: P(s) + Cl2(g) PCl5(s) Balance this equation and determine how many moles of chlorine are consumed if 5.50 grams of PCl5 are produced.

26. 2 P 2 10 Cl 10 5.50 g PCl5 1 mol PCl5 208 g PCl5 5 mol Cl2 2 mol PCl5
2P(s) + 5Cl2(g) PCl5(s)
2 P 2
10 Cl 10
5.50 g PCl5
1 mol PCl5
208 g PCl5
5 mol Cl2
2 mol PCl5
_mol Cl2