Calculations

Question 1

(1) 1.27 g of copper were produced in an experiment.
Calculate the number of moles of copper, Cu, produced in this experiment.
(Relative atomic mass: Cu = 63.5

Answer 1

1.27 / 63.5 (1) (= 0.02)

Question 2

(1) Calculate the minimum volume of hydrogen required to completely convert
1000 dm3 of nitrogen into ammonia.

Answer 2

3 x 1000 (1) (= 3000)

Question 3

(3) NH3 + HNO3 → NH4NO3
Calculate the mass of ammonium nitrate produced by the complete reaction
of 34 g of ammonia
(Relative atomic masses H = 1.0, N = 14, O = 16)

Answer 3

Method 1
14 + (3 x 1) (1) g of NH3
makes 14 + (4 x 1) + 14 + (3 x 16) (1) g
NH4NO3

34 g of NH3 makes
(80 x 34)/ 17

Question 4

(1) volume of sodium hydroxide solution = 25.0 cm3
rough titration = 23.1 cm3
1st titration = 22.6 cm3
2nd titration = 22.8 cm3
State the volume of hydrochloric acid that must be used to calculate the
concentration of sodium hydroxide solution.

Answer 4

(22.6 + 22.8)/ 2 (1) (= 22.7)

Question 5

(3) In a different experiment, 25.0 cm3
of sodium hydroxide solution reacted with
23.2 cm3
of 0.100 mol dm–3 hydrochloric acid, HCl.
Calculate the concentration of this sodium hydroxide solution, NaOH, in
moldm–3.

Answer 5

use of c1v1 = c2v2 (1)
0.1x 23.2 = conc x 25.0 (1)
conc NaOH = (0.1 x 23.2)/25.0 (1)
(= 0.0928) mol dm-3

Question 6

(2) A sodium hydroxide solution was made up by dissolving 20.0 g of sodium
hydroxide in water and making the volume of the solution up to 1.00 dm3
.
Calculate the concentration of sodium hydroxide, NaOH, in this solution
in mol dm–3.
(relative atomic masses: H = 1.00, O = 16.0, Na = 23.0)

Answer 6

rel mass NaOH = 23.0 + 16.0 +
1.00 (1)

concentration = 20.0 x 1 (1)
formula mass

Question 7

(3) In another experiment, a titration was carried out.
25.0 cm3
of 1.50 mol dm–3 sodium hydroxide solution, NaOH, was titrated with
hydrochloric acid.
The volume of the hydrochloric acid required to neutralise the sodium hydroxide
solution was 30.0 cm3
.
Calculate the concentration of the hydrochloric acid, HCl, in mol dm–3.
HCl + NaOH ĺ NaCl + H2
O

Answer 7

moles of NaOH = (25.0 x 1.50) /1000  (1)
 (= 0.0375 moles)
ratio 1 : 1 /
moles NaOH = moles HCl (1)
conc of HCl = (0.0375 x1000) /30.0 (1)
 (= 1.25 (mol dm-3))
OR
25.0 x 1.50 = 30.0 x conc acid (2)
conc of HCl = (0.0375 x1000) /30.0 (1)
 (=1.25 (mol dm-3))

Question 8

(1) If 200 cm3
of hydrogen react completely with 100 cm3
of oxygen, what is the
maximum volume of water vapour formed, if all volumes are measured at the
same temperature and pressure?

Answer 8

200 cm3

Question 9

Calculate the maximum volume of hydrogen formed, at room temperature and
pressure, when 13.0 g of zinc reacts completely with excess hydrochloric acid.
(relative atomic mass: Zn = 65.0,
1 mol of any gas occupies 24 dm3
at room temperature and pressure)

Answer 9

65.0 g Zn produces 24 dm3
H2 (1)
13.0 g Zn produces (13.0 x 24)/ 65.0(1)
(= 4.8 dm3 H2)

Question 10

(1) Calculate the minimum volume of nitrogen, in dm3
, required to react
completely with 1000 dm3
of hydrogen.

Answer 10

333 dm3

Question 11

(3) 11.1g of calcium chloride are dissolved in water.
The volume of the solution is made up to 500 cm3
.
Calculate the concentration, in mol dm–3, of calcium chloride, CaCl2
, in this
solution.
(relative atomic masses: Cl = 35.5, Ca = 40.0)

Answer 11

CaCl2 = 40 + 35.5 + 35.5 (=111)
(1)
THEN
moles = 11.1 / 111 (= 0.1)
(1)
conc = moles x 1000/500 (=0.2)
(1)
OR
mass conc = 11.1 x 1000/500 (=22.2)
(1)
conc = mass conc/111 (= 0.2)
(1)

Question 12

(1) The burette readings of acid added were
titration 1 titration 2 titration 3
final volume / cm3 27.20 30.10 25.35
initial volume / cm3 2.05 5.20 0.10
volume of acid added / cm3 25.15 24.90 25.25
The volume of acid added that should be used in the calculation is

Answer 12

25.20 cm3