Question

what mass of iron sulfide is obtained if 8 grams of sulfur and 28 grams of iron are taken for the reaction?​

Answer 1

Mass of iron sulfide obtained : 22 g

Further explanation

Reaction

Fe+S⇒FeS

8 grams of sulfur and 28 grams of iron are taken for the reaction

• Find limiting reactant

mol of Fe(Ar=56 g/mol) :

`\tt mol=(mass)/(Ar)\\\\mol=(28)/(56)\\\\mol=0.5`

mol of S(Ar=32 g/mol) :

`\tt mol=(8)/(32)=0.25`

Limiting reactant ⇒ S(smaller mol ratio), excess reactant : Fe

mol of FeS based on limiting (S) ⇒0.25(mol ratio from equation = 1 : 1)

Mass FeS(MW=88 g/mol) :

`\tt mass=mol* MW\\\\mass=0.25* 88\\\\mass=22~g`