Answer:
The mass of iron sulphide formed is 14.2 g
Step-by-step explanation:
Starting off with a balanced chemical equation:
2Fe + 3S --> Fe₂S₃
to find the mass of Fe₂S₃ and thus the limiting reagent, we can first find the number of moles of each reactant. Number of moles is found by dividing mass by molar mass (found using a standard IUPAC Periodic Table).
n(Fe) present = m/M = 7.62/55.85 = 0.13644 mol
n(S) present = m/M = 8.67/32.07 = 0.27035 mol.
Using the values for the number of moles of each reactant, we can find the limiting reagent by testing each with the molar ratio of the equation.
If (Fe) is the limiting reagent, then moles of sulphur required = n(Fe)/2 × 3
= 0.20466 mol. Since we have excess of what's required of sulphur, then iron could be the limiting reagent.
Now we test it with (S). If (S) is the limiting reagent, then moles of iron required = n(S)/3 × 2 = 0.18023 mol. Since we have less than what is required, therefore, sulphur is NOT the limiting reagent.
Hence, Iron is the limiting reagent.
Now we have the limiting reagent, we can use this to calculate the number of moles of Fe₂S₃, since the limiting reagent is the reactant that is completely consumed.
Since stoichiometry of Fe : Fe₂S₃ = 2 : 1, therefore:
moles of (Fe) = 2× moles of (Fe₂S₃).
hence, moles of (Fe₂S₃) = 1/2 × 0.13644 mol = 0.06822.
Finally, we can multiply the moles of iron sulphide by the molar mass, to get the mass.
m(Fe₂S₃) = nM = 0.06822 × (55.85×2 + 32.07×3) = 14.2 g (3 significant figures, since all the data in the question is in 3 sig. figs).