Question

Calculate the number of moles of $F{e}_2{O}_3$ when 600 g of $Fe{S}_2$, reacts with 800 g of $O_2.$
$4Fe{S}_2+11O_2\to 2F{e}_2{O}_3+8S{O}_2$
(Given: Molar mass of Fe is 56 g/mol)

• A. 1 mol
• B. 1.5 mol
• C. 2 mol
• D. 2.5 mol

Answer 1

The correct option is D 2.5 mol

Given balanced chemical reaction is,
$4Fe{S}_2+11O_2\to 2F{e}_2{O}_3+8S{O}_2$
We know,
Molar mass of $Fe{S}_2=(56+32\times 2)g/mol=120g/mol$

Molar mass of $O_2$ = 32 g/mol
$\therefore 600gofFe{S}_2=\frac{600}{120}$ mol $Fe{S}_2=5molofFe{S}_2$

800 g $O_2=\frac{800}{32}molof{O}_2=25molof{O}_2$

From the balanced reaction,
11 moles of $O_2$ reacts with 4 moles of $Fe{S}_2$
Therefore, 25 moles of $O_2$ will react with
$=\frac{4}{11}\times 25molFe{S}_2=9.09molFe{S}_2$

So, $Fe{S}_2$ is the limiting reagent here.

Therefore, moles of $F{e}_2{O}_3$ produced
$=\frac{2}{4}\times 5$ mol = 2.5 mol.