Question

Given the Law of Conservation of Matter, if 3.56g of Fe reacts completely with 1.53g of O2, what mass of Fe2O3 should be produced?

4Fe(s) + 3O2(g) --> 2Fe2O3(s)

Answer 1

Molar mass of Fe = 55.85 g/mol


mass(Fe)= 3.56 g

number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(3.56 g)/(55.85 g/mol)
= 6.374*10^-2 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 1.53 g

number of mol of O2,
n = mass of O2/molar mass of O2
=(1.53 g)/(32 g/mol)
= 4.781*10^-2 mol
Balanced chemical equation is:
4 Fe + 3 O2 ---> 2 Fe2O3


4 mol of Fe reacts with 3 mol of O2
for 0.063742 mol of Fe, 0.047807 mol of O2 is required
But we have 0.047813 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


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 = (2/3)* moles of O2
= (2/3)*0.047813
= 0.031875 mol


mass of Fe2O3 = number of mol * molar mass
= 3.188*10^-2*1.597*10^2
= 5.09 g

Answer: 5.09 g