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

Mass of iron sulfide obtained : 22 g

Reaction

Fe+S⇒FeS

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

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

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

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) :

43.452

step-by-step explanation:

284 times .153 = 43.452

i just learnt this recently but i need to look through my notes again