Question

 Iron reacts with oxygen at high temperatures to form iron(III) oxide

Answer 1

1)
Molar mass of Fe = 55.85 g/mol


mass(Fe)= 15.7 g

use:
number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(15.7 g)/(55.85 g/mol)
= 0.2811 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 20.3 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(20.3 g)/(32 g/mol)
= 0.6344 mol
Balanced chemical equation is:
4 Fe + 3 O2 ---> 2 Fe2O3 +


4 mol of Fe reacts with 3 mol of O2
for 0.2811 mol of Fe, 0.2108 mol of O2 is required
But we have 0.6344 mol of O2

so, Fe is limiting reagent
Answer: Fe

2)
Molar mass of Fe2O3,
MM = 2*MM(Fe) + 3*MM(O)
= 2*55.85 + 3*16.0
= 159.7 g/mol

According to balanced equation
mol of Fe2O3 formed = (2/4)* moles of Fe
= (2/4)*0.2811
= 0.1406 mol


use:
mass of Fe2O3 = number of mol * molar mass
= 0.1406*1.597*10^2
= 22.45 g
Answer: 22.5 g

3)
% yield = actual mass*100/theoretical mass
= 7.6*100/22.45
= 33.86%
Answer: 33.9 %