What mass of fes is formed if 9.42 g of fe reacts with 68.0 g of s8? 8fe(s) + s8(s) → 8fes(s)?

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.

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