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Mole Calculations Assignment Two – Answers
1.
Sodium reacts with water to form an aqueous solution of sodium hydroxide and hydrogen.
a)
Write the balanced chemical equation, including state symbols, for this reaction:
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
b) A student added 34.5 g of sodium to 250 cm3 of distilled water.
i)
Calculate the volume of hydrogen gas produced by this reaction.
mol of Na = mass in g  Ar
mol of Na = 34.5  23 = 1.50 mol
from the balanced chemical equation, 2 mol of Na produce 1 mol of H2
 1.50 mol of Na will produce 1/2  1.50 = 0.750 mol of H2
volume of H2 in dm3 = mol  24
volume of H2 in dm3 = 0.750  24 = 18.0 dm3
ii)
Calculate the concentration of sodium hydroxide produced by this reaction.
from the balanced chemical equation, 2 mol of Na produce 2 mol of NaOH
 1.50 mol of Na will produce 2/2  1.50 = 1.50 mol of NaOH
there are 1.50 mol of NaOH dissolved in 250 cm3 of solution
concentration of NaOH = (1.50  250)  1000 = 6.00 mol/dm3
2.
Ammonia burns in oxygen to produce nitrogen and water vapour.
a)
Write the balanced chemical equation, including state symbols, for this reaction.
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
b) Calculate the volume of water vapour that is produced when 42.5 g of ammonia burn
completely in oxygen.
mol of NH3 = mass in g  Mr
mol of NH3 = 42.5  (14 + 1 + 1 + 1) = 42.5  17 = 2.50 mol
from the balanced chemical equation, 4 mol of NH3 produce 6 mol of H2O
 2.50 mol of NH3 will produce 6/4  2.50 = 3.75 mol of H2O
volume of H2O in dm3 = mol  24
volume of H2O in dm3 = 3.75  24 = 90.0 dm3
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)
3.
Aspirin can be made from 2-hydroxybenzoic acid according to the following chemical equation:
O
H
H
C
C
C
C
C
O
C
C
H
O
H
H
H
2-Hydroxybenzoic
Acid
a)
O
O
H3C C
+
H2SO4
H
O
(catalyst)
H3C C
H
O
C
C
C
C
C
O
C
C
H
H
O
C
H
C
H
H
H
+
H C C
O
H
2-Ethanoylhydroxybenzoic
Acid (Aspirin)
Ethanoic
Anhydride
O
H
O H
Ethanoic
Acid
What mass of aspirin can be made from 759 g of 2-hydroxybenzoic acid?
formula of 2-hydroxybenzoic acid = C7H6O3
Mr 2-hydroxybenzoic acid = (7  12) + (6  1) + (3  16) = 138
mol 2-hydroxybenzoic acid = mass in g  Mr
mol 2-hydroxybenzoic acid = 759  138 = 5.50 mol
from the balanced equation, 1 mol of 2-hydroxybenzoic acid produces 1 mol of aspirin
 5.50 mol of 2-hydroxybenzoic acid produces 1/1  5.50 = 5.50 mol of aspirin
formula of aspirin = C9H8O4
Mr aspirin = (9  12) + (8  1) + (4  16) = 180
mass of aspirin in g = mol  Mr
mass of aspirin in g = 5.50  180 = 990 g
b) The reaction only produced 960.3 g of aspirin. Calculate the percentage yield.
percentage yield of aspirin = (experimental yield  theoretical yield)  100
percentage yield of aspirin = (960.3  990)  100
percentage yield of aspirin = 97.0 %
4.
A rock sample contains a mixture of magnesium carbonate and sand. 90.0 g of the rock
sample was found to react exactly with 135 cm3 of 8.00 mol/dm3 hydrochloric acid.
a)
Write the balanced chemical equation, including state symbols, for the reaction between
magnesium carbonate and hydrochloric acid.
MgCO3(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) + CO2(g)
b) Using the information given, calculate what percentage of the rock sample is magnesium
carbonate.
moles of hydrochloric acid used = c  v  10–3
moles of hydrochloric acid used = 8.00  135  10–3 = 1.08 mol
from the balanced chemical equation, 2 mol of HCl react with 1 mol of MgCO3
 1.08 mol HCl reacts with 1/2  1.08 = 0.540 mol of MgCO3
mass of MgCO3 in grams = mol  Mr
mass of MgCO3 in grams = 0.540  (24 + 12 + (3  16)) = 0.540  84 = 45.36 g
percentage MgCO3 in the rock by mass = (45.36  90.0)  100 = 50.4 %
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5.
An aqueous solution of an unknown metal nitrate, MNO3, was reacted with an excess of
aqueous sodium hydroxide. A white precipitate of the metal hydroxide, MOH, was formed. The
balanced chemical equation for the reaction is given below:
MNO3(aq) + NaOH(aq)  MOH(s) + NaNO3(aq)
40.0 cm3 of 5.00 mol/dm3 aqueous sodium hydroxide were found to precipitate 25.0 g of the
metal hydroxide. Use this information to calculate the relative molecular mass of metal M, and
hence identify the metal.
moles of NaOH used = c  v  10–3
moles of NaOH used = 5.00  40.0  10–3 = 0.200 mol
from the balanced chemical equation, 1 mol of NaOH produces 1 mol of MOH
 0.200 mol of NaOH produces 1/1  0.200 = 0.200 mol of MOH
Mr of MOH = mass in grams  mol
Mr of MOH = 25.0  0.200 = 125
Ar of M = 125 – Ar[O] – Ar[H]
Ar of M = 125 – 16 – 1 = 108
the element with Ar = 108 is silver
further confirmation, the formula of silver nitrate is AgNO3 and silver hydroxide is AgOH
6.
Acids react with alkalis to form a salt and water.
a)
Write the balanced chemical equation for the reaction between sulfuric acid and aqueous
sodium hydroxide.
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l)
b) 150 cm3 of 0.200 mol/dm3 aqueous sodium hydroxide were reacted with 50.0 cm3 of
0.400 mol/dm3 sulfuric acid. What is the limiting reagent for this reaction?
moles of NaOH used = c  v  10–3
moles of NaOH used = 0.200  150  10–3 = 0.0300 mol
moles of H2SO4 used = c  v  10–3
moles of H2SO4 used = 0.400  50.0  10–3 = 0.0200 mol
from the balanced chemical equation, 1 mol of H2SO4 react with 2 mol of NaOH
 0.0200 mol H2SO4 react with 2/1  0.0200 = 0.0400 mol of NaOH
however, only 0.0300 mol of NaOH are available, therefore NaOH is the limiting reagent
OR
from the balanced chemical equation, 2 mol of NaOH react with 1 mol of H2SO4
 0.0300 mol NaOH react with 1/2  0.0300 = 0.0150 mol of H2SO4
0.0300 mol of H2SO4 are available, therefore H2SO4 is present in excess and NaOH is the
limiting reagent
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