Question

how much fes would form if 5.6 grams of fe reacted with 3.2 grams of s?

reaction	reactant(s)	product(s)
mass	5.6 g	3.2 g	?

Explanation:

Step1: Calculate moles of Fe and S

Molar mass of Fe is approximately 56 g/mol, so moles of Fe = $\frac{5.6\ g}{56\ g/mol}= 0.1\ mol$. Molar mass of S is approximately 32 g/mol, so moles of S = $\frac{3.2\ g}{32\ g/mol}=0.1\ mol$.

Step2: Determine the limiting reactant

The chemical equation for the reaction is $Fe + S
ightarrow FeS$. The mole - ratio of Fe to S to FeS is 1:1:1. Since the moles of Fe and S are equal (0.1 mol each), neither is in excess and both react completely.

Step3: Calculate mass of FeS formed

Molar mass of FeS = 56 + 32=88 g/mol. Since 0.1 mol of Fe reacts completely and 1 mol of Fe forms 1 mol of FeS, 0.1 mol of FeS is formed. Mass of FeS = 0.1 mol×88 g/mol = 8.8 g.

Answer:

8.8 g