equations

Question 1

Nitrogen and hydrogen form ammonia shown by the following
equation:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
Calculate the mass of nitrogen needed to form 6.8 tonnes of
ammonia.
Relative atomic masses (A
r): H = 1; N = 14

Answer 1

Step 1 - Work out the number of number of moles of ammonia (Mr of ammonia = 17)
6800000 / 17 = 400000 moles of ammonia
Step 2 - Use the balanced equation and number of moles of ammonia to work out the number of moles of
nitrogen
The ratio of nitrogen to ammonia is 1:2
Therefore the number of moles of nitrogen is 400000/2 = 200000
Step 3 - Work out the mass of nitrogen (Mr of N2 is 28)
200000 x 28 = 5600000 g = 5.6 tonnes.

Question 2

Hydrogen peroxide decomposes in
water to form water and oxygen. How
many grams of oxygen gas will be
given off from 40.8 g of hydrogen
peroxide?

Answer 2

Step 1: Write the balanced equation 2 H2O2(l) → 2 H2O + O2(g) Mr of H2O2 = 34
Step 2: Number of moles in 40.8 g : 40.8/34 = 1.2 moles
Ratio in the balanced equation of H2O2 : O2 = 2:1
Step 3 :Therefore number of moles of O2 = 0.6 moles
Step 4: Mass of oxygen = 0.6 x 32 (Mr of O2) = 19.2

Question 3

31.0 cm3 of potassium hydroxide solution
neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.
HNO3 + KOH → KNO3 + H2O
Calculate the concentration of the potassium
hydroxide solution in moldm−3

Answer 3

Calculate the concentration of the potassium
hydroxide solution in moldm−3
Step 1: Calculate the moles of HNO3 used = Concentration x volume
2 x 0.025 dm3 (25/1000 to convert the units) = 0.05 moles
Step 2 : Calculate the moles of KOH
Ratio is 1:1 therefore number of moles of KOH = 0.05
Step 3 : Calculate the concentration of KOH
Volume = Moles/concentration; 0.05 / 0.031 = 1.61

Question 4

What is the % yield of NH3 if 40.5 g
NH3 is produced from 20.0 mol H2
and excess N2?

Answer 4

Step 1 - Write the balanced equation
N2 + 3 H2 → 2 NH3
Step 2 - Calculate the theoretical amount of NH3 .Moles NH3 (ratio of H2 to NH3 is 3:2);
of 20/1.5 = 13.3 moles
13.3 X 17 (Mr of NH3) = 227
Step 3 - Calculate percentage yield of NH3
40.5/227 x 100 = 17.8%

Question 5

Look at the equations for the two reactions that produce CuCl2
Reaction I: CuCO3(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)
Reaction II: CuO(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l)
Reactive formula masses: CuO = 79.5; HCl = 36.5; CuCl2 = 134.5;
H2O = 18
Which reaction has a better atom economy?

Answer 5

Reaction II (look at the reactants):
Total formula mass of reactants = 152.5
Formula mass of CuCl2 = 134.5
(134.5/152.5) x 100% = 88.2%

Question 6

What is the relative formula
mass of:
A) CaF2
B) B) C6H12O6

Answer 6

CaF2 - (Ar values: Ca = 40, F = 19)
40 + 19 + 19 = 78
---
C2H12O6 - (Ar values: C = 12, H = 1, O = 16)
(12 x 6) + (1 x 12) + (16 x 6) = 180

Question 7

What is the mass of:
20 moles of calcium
carbonate, CaCO3

Answer 7

Mass = Mr x Moles
Mr = 100
100 x 20 = 2000 g

Question 8

Calculate the amount of carbon dioxide
in moles in 0.32 g of carbon dioxide.
Relative atomic masses (A
r): carbon =
12, oxygen = 16

Answer 8

Moles = Mass / Mr

0.32 / 44 = 0.007