From the equation, we see that the molar ratio of Fe : S required is:
8 : 1
The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265
The molar ratio is:
1 : 1.6
Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of
Fe : FeS is
1 : 1
So 0.168 moles of FeS will form. The mass of FeS will be:
Mass = 0.168 * (56 + 32)
Mass = 14.78 grams
14.78 grams of FeS will be formed.