Chemistry 112
Chapter 10 Exercise: Review Problems
Answers
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1. Show your work to find the molar mass of calcium phosphate.
Ca3(PO4)2 = 3 Ca + 2 P + 8 0
= 3(40.08) + 2(30.97) + 8(16.00)
= 310.18
2. How many grams in 0.38 moles of calcium phosphate?
0.38 mol 310.18 g = 117.8684 g = 120 g (2 SDs)
‰ mol 
3. How many moles in 5.87 g calcium phosphate?
5.87 g mol
= 0.0189244 mol = 0.0189 mol (3 SDs)
‰310.18 g
or 1.89 x 10-2 mol
4. How many formula units (particles) in 10.50 moles of calcium
phosphate?
10.50 mol 6.02 x 1023 form.un. = 6.32 x 1024 form. un (3 SDs)
‰
mol

25
5. How many kg in 3.855 x 10 formula units of calcium phosphate?
3.855 x 1025 form. un.
mol
310.18 g kg
23
‰6.02 x 10 form.un.‰ mol ‰103 g
= 19.9 kg (3 SDs)
6. How many moles in 45 L of carbon dioxide at STP?
45 L mol
= 2.0089 mol = 2.0 mol (2 SDs)
‰22.4 L
7. What volume would be occupied by 5.2 mol CO2(g) at STP?
5.2 mol 22.4 L = 116.48 L = 120 L (2 SDs)
‰ mol 
8. What would be the mass of 100.0 L CO2(g) at STP?
100.0 L mol 44.01 g = 196.4732 g = 196 g (3 SDs)
‰22.4 L‰ mol 
9. How many grams in 1.92 mol ammonium phosphate? (mol—>g)
1.92 mol 149.12 g = 286 g (286.3104 with 2 SDs)
‰ mol 
10. How many particles (form. un.) would there be in the previous
question? (mol —> particles)
2.1.92 mol 6.02 x 1023 molecules = 1.16 x 1024 molecules
‰
mol
 (1.15584 x 1024 with 3 SDs)
11. How many litres of oxygen (O2(g))would be occupied by 5.25 moles
at STP? (mol —> L)
Chemistry 112
Chapter 10 Exercise: Review Problems
Answers
5.25 mol 22.4 L = 118 L (117.6 L with 3 SDs)
‰ mol 
12. 15.50 L oxygen would contain how many grams at SATP?
(L—>mol—>g)
15.50 L mol 32.00 g = 20.0 g = (20 with 3 SDs)
‰24.8 L‰ mol 
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