Fe(s) + S(s) → FeS(s) In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S.
What is the limiting reagent, and what is the reactant in excess?
Calculate the mass of FeS formed.

The mass of FeS formed : 11.97 g

Reaction

Fe(s) + S(s) → FeS(s)

mass Fe=7.62 g

mol Fe(MW=56 g/mol) :

mass S=8.67 g

mol S(MW=32 g/mol) :

Limiting reactant : smaller ratio(mol:coefficient) ⇒coefficient=1

Fe : S =0.136 : 0.271⇒Fe limiting(smaller), S excess

Mol FeS based on Fe , so mol FeS=0.136

Mass FeS(MW=88 g/mol) :

1.solid

2.carbon dioxide

3.helium

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2