Question

What mass of iron (II) sulfide, FeS (MM=87.92g/mol) is formed when 5.6g of iron, Fe (MM=55.85g/mol) is completely reacted with excess sulfur? Use the reaction;
Fe + S -> FeS

Answer 1

Fe + S -> FeS

Moles of Fe $=\frac{5.6}{55.85}=0.1moles$

Mass of FeS$=87.92×0.1=8.792grams$