Practice: Moles and Stoichiometry
1. Ans: A
m = 8.5 g of NaNO3
M = (22.99 + 14.01 + 3x16.00) = 85g/mol of NaNO3
n = m/M
n = 8.5g/85g/mol
n = 0.1 mol of NaNO3
molar ratio:
1 molecule of NaNO3 for every 1 molecule of HNO3
so 0.1mol of NaNO3 : 0.1mol of HNO3
n = 0.1 mol of HNO3
M = (1.01 + 14.01 + 3x16.00) = 63.02g/mol of HNO3
m=nxM
m = 0.1mol x 63.02g/mol
m = 6.3g of HNO3
2. Ans: B
3. Ans: 31.43g/mol of AgNO3
m = 2 g of Ag
M = 107.87g/mol of Ag
n = m/M
n = 2g/107.87g/mol
n = 0.0185 mol of Ag
molar ratio:
2 molecule of Ag for every 2 molecule of AgNO3
so 0.0185mol of Ag : 0.0185mol of AgNO3
n = 0.0185 mol of AgNO3
M = (107.87 + 14.01 + 3x16.00) = 169.88g/mol of AgNO3
m=nxM
m = 0.0185mol x 169.88g/mol
m = 31.4278g of AgNO3
4. Ans: C
n =m/M
m = 10g
M = (22.99 + 16.00 + 1.01) = 40g/mol of NaOH
n = 10g / 40g/mol = 0.25 moles of NaOH
m = 73g
M = (22.99 + 35.45) = 58.44g/mol of NaCl
n = 73g / 58.44 g/mol = 1.25 moles of NaCl
m = 126g
M = (1.01 + 14.01 + 3x16.00) = 63.02g/mol of HNO3
n = 126g / 63.02 g/mol = 2.00 moles of HNO3
m = 140g
M = (200.59 + 79.91) = 280.5g/mol of HgBr
n = 140g / 280.5 g/mol = 0.5 moles of HgBr
5. Ans: 116.66g of Mg(OH)2
Molar Ratio: 2 molecules of HCl for every 1 molecule of Mg(OH)2
So
4 moles of HCL : 2 moles of Mg(OH)2
n = 2 moles of Mg(OH)2
M = (24.31 + 2x16.00 + 2x1.01) = 58.33g/mol of Mg(OH)2
m=nxM
m = 2 mol x 58.33g/mol
m = 116.66g of Mg(OH)2
6. Ans: 27.23g of CaSO4
C = n/V
C = 1.0mol/L of Na2SO4
V = 200ml = 0.2L
n=CxV
n = 1.0mol/L x 0.2L
n = 0.2 moles of Na2SO4
Molar Ratio: 1 molecule of Na2SO4 for every 1 molecule of CaSO4
So
0.2 mol of Na2SO4 : 0.2 mol of CaSO4
n = 0.2 moles of CaSO4
M = (40.08 + 32.06 + 4x16) = 136.14g/mol of CaSO4
m=nxM
m = 0.2 mol x 136.14 g/mol
m = 27. 23 g of CaSO4
7. Ans: E
** We did not learn this in class, so you do not need to complete it however, here is how
you would do the question.
Soft drink pH = 3
pOH = 14 – 3 = 11
3 = - log [H+]
-1
-1
11 = - log [OH-]
-1
-1
3 = log [H+]
11 = log [OH-]
10-3 = [H+]
10-11 = [OH-]
8. Ans: C
m = 80g of CH4
M = (4x1.01 + 12.01) = 16.05g/mol
n = m/M
n = 80g/16.05g/mol
n = 4.984 moles of CH4 ~ 5 moles
9. Ans: 16.65g of CaCl2
n = 0.15 moles of CaCl2
M = (40.08 + 2x35.45) = 110.98 g/mol
m=nxM
m = 0.15mol x 110.98g/mol
m = 16.647 g of CaCl2
10. Ans: 16.8kg of NaOH
m = 924g of CO2/day
M = (12.01 + 2x16.00) = 44.01g/mol
n = m/M
n = 924g/44.01g/mol
n = 20.995 moles of CO2
Molar Ratio: For every 1 molecule of CO2 we get 2 molecules of NaOH
So
20.995 moles of CO2 : 41.99 moles of NaOH
n = 41.99 moles
M = (22.99 + 16.00 + 1.01) = 40g/mol
m=nxM
m = 41.99mol x 40g/mol
m = 1679.6g of NaOH /day
Amount for ten days = 1679.6g/day x 10days = 16.8 kg of NaOH
11. Ans: D
** We have not gone over properties of states of matter yet; however the substance that
occupies the most volume is the substance in the gaseous state. The molecules in a gas are
more spaced out and therefore occupy a greater volume with the same number of moles as
a liquid or a solid. This means that 70g of chlorine (g) will have the greatest volume.
12. Ans: C
13. Ans: A
3CuO + 2NH3  N2 + 3Cu + 3H2O
n = 9 mol of Cu
Molar Ratio: For every 3 molecules of Cu you have 2 molecules of NH3
So
3 : 2
9 mol : X
(9)(2) = 3(X)
(9)(2)/(3) = X
X = 6 moles of NH3
Cross multiply