Calculate the mass of the coal sample in kg containing 0.05% mass of iron pyrite fes2 that can produced 44.8L of so2 at 1atmosphere and 273k with 40% reaction yield

Dear Student,
The reaction will be
$Fe{S}_2+\frac{5}{2}{O}_2\underset{}{\to }FeO+2S{O}_2$
44.8L of SO₂ is produced at 1 atm and 273K
No. of moles of SO₂ = $\frac{44.8}{22.4}=2 mol$
But the reaction yield is 40%
2 mol of SO₂ is produced if the yield is 100% from 1 mol FeS₂
2 mol of SO₂ is produced if the yield is 40% from $\frac{1 x 100}{40}=\frac{5}{2}mol$ FeS₂
Mass of FeS₂ = $\frac{5}{2}x120 =300g [Molar mass of Fe{S}_2=120 g/mol]$
The sample of coal contains 0.05% FeS₂ by mass
300 g FeS₂ is present in $\frac{300 x 100}{0.05}=600000 g = 600kg$
Mass of coal = 600kg

Regards