To determine the mass of iron needed to react with 32.0 grams of sulfur to produce iron (II) sulfide, we first need to balance the chemical equation for the reaction:
\[ \text{Fe} + \text{S}_8 \rightarrow \text{FeS} \]
Now, we need to find the molar masses of iron (Fe) and sulfur (S) from the periodic table:
- The molar mass of iron (Fe) is approximately 55.85 g/mol.
- The molar mass of sulfur (S) is approximately 32.06 g/mol.
Next, we need to determine the stoichiometry of the reaction. From the balanced equation, we see that 1 mole of iron reacts with 1 mole of sulfur to produce 1 mole of iron (II) sulfide.
Now, let's calculate the number of moles of sulfur in 32.0 grams:
\[ \text{moles of sulfur} = \frac{\text{mass}}{\text{molar mass}} = \frac{32.0 \, \text{g}}{32.06 \, \text{g/mol}} \approx 0.998 \, \text{mol} \]
Since the reaction is 1:1 between iron and sulfur, the number of moles of iron required will also be approximately 0.998 moles.
Finally, we can calculate the mass of iron needed:
\[ \text{mass of iron} = \text{moles} \times \text{molar mass} = 0.998 \, \text{mol} \times 55.85 \, \text{g/mol} \approx 55.78 \, \text{g} \]
So, approximately 55.78 grams of iron are needed to react with 32.0 grams of sulfur.