Unit 1 worksheets 3 Reacting masses and volumes
1. Zinc metal reacts with copper(II) sulfate solution according to the following equation:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Determine the maximum mass of copper that can be deposited when 1.20 g of zinc is added to
50.0 cm3 of 2.00 x 10-1 mol dm-3 copper(II) sulfate solution.
2. Calculate the mass of carbon dioxide produced when 150 cm3 of 1.00 mol dm-3 hydrochloric acid,
HCl(aq), is added to 10.0 g of calcium carbonate, CaCO3.
3. A student prepared ethene by dehydrating ethanol.
C2H5OH(l) → C2H4(g) + H2O(l)
She started with 9.36 g of ethanol and made 2.12 g of ethene. Calculate the percentage yield she
obtained for this reaction.
4. A sample of gas occupies 67.2 cm3 at a temperature of 22 oC and a pressure of 9.38 x 104 Pa.
Calculate the volume the gas will occupy if the temperature is increased to 29 oC and the pressure
increased to 1.06 x 105 Pa.
5. 2.50 dm3 of gas at a temperature of 19 oC and a pressure of 1.01 x 105 Pa has a mass of 4.59 g.
Determine the molar mass of the gas.
6. The molecular formula of a gaseous hydrocarbon can be determined by combusting it completely
in excess oxygen and then passing it through potassium hydroxide solution to absorb the carbon
dioxide produced. In an experiment 200 cm3 of a hydrocarbon was reacted with 1500 cm3 of
oxygen. After the hydrocarbon had combusted completely 1000 cm3 of gas remained. This volume
was reduced to 200 cm3 after the gas had been passed through a solution of potassium hydroxide.
All volumes were measured under the same conditions of temperature and pressure.
Deduce the formula of the hydrocarbon.
Answers
1. Amount of zinc = 1.20/65.38 = 1.84 x 10-2 mol
Amount of copper(II) sulfate = 50/1000 x 2.00 x 10-1 = 1.00 x 10-2 mol
Zinc is in excess and copper(II) sulfate is the limiting reagent
Maximum amount of copper deposited = 1.00 x 10-2 mol
Mass of copper deposited = 1.00 x 10-2 x 63.55 = 6.36 x 10-1 g (0.636 g)
2. Amount of hydrochloric acid = 150/1000 x 1.00 = 1.50 x 10-1 mol
Amount of calcium carbonate = 10.0/[40.08 + 12.01 + (3 x 16.00)] = 1.00 x 10 -1 mol
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
Calcium carbonate is in excess and the hydrochloric acid is the limiting reagent
Amount of carbon dioxide produced = ½ x 1.50 x 10-1 mol = 7.50 x 10-2 mol
Mass of carbon dioxide produced = 7.50 x 10-2 x 44.01 = 3.30 g
3. Amount of ethanol = 9.36/[(2 x 12.01) + (6 x 1.01) + 16.00] = 2.03 x 10 -1 mol
Maximum amount of ethene that could be formed = 2.03 x 10-1 mol
Maximum mass of ethene that could be formed = 2.03 x 10-1 x [(2 x 12.01) + (4 x 1.01)] = 5.696 g
Percentage yield = (2.12/5.696) x 100 = 37.2 %
4. Volume = 67.2 x 302 x 9.38 x 104 = 60.9 cm3
295 1.06 x 105
5. n = PV/RT
n = (1.01 x 105) x (2.50 x 10-3)/ 8.31 x 292 = 1.04 x 10-1 mol
Molar mass of gas = 4.59/1.04 x 10-1 = 44.1 g mol-1
6. All hydrocarbons (represented by CxHy) combust completely to give carbon dioxide and water
CxHy(g) + (x + y/4)O2(g) → xCO2(g) + y/2H2O(l)
After the reaction the volume of the carbon dioxide produced and the excess oxygen = 1000 cm3
Volume of carbon dioxide = 800 cm3 so volume of excess oxygen = 200 cm3
200 cm3 of CxHy reacts with 1300 cm3 of O2 to produce 800 cm3 of CO2
I volume of CxHy reacts with 6.5 volumes of O2 to produce 4 volumes of CO2
Equal volumes of different gases under the same conditions contain the same number of particles
x = 4 and (x + y/4) = 6.5 so y = 10
The molecular formula of the hydrocarbon is C4H10