Final answer:
To find the volume of 0.200 m FeCl₃ needed to react and produce 1.38 g of Fe₂S₃ with 65.0% yield, we calculate the number of moles of Fe₂S₃, adjust for yield, find the moles of FeCl₃ using the mole ratio from the chemical equation, and then convert to volume. The result is that 41.46 mL of FeCl₃ is required.
Step-by-step explanation:
To calculate how many milliliters of 0.200 m FeCl₃ are needed to react with an excess of Li₂S to produce 1.38 g of Fe₂S₃ with a percent yield of 65.0%, we need to work backwards from the given mass of Fe₂S₃ to the volume of FeCl₃.
First, we convert the mass of Fe₂S₃ to moles:
Fe₂S₃ has a molar mass of approximately 207.87 g/mol for iron (Fe) and 32.07 g/mol for sulfur (S), totaling to 207.87 * 2 + 32.07 * 3 = 511.95 g/mol for Fe₂S₃. Starting with 1.38 g:
1.38 g Fe₂S₃ * (1 mol Fe₂S₃ / 511.95 g Fe₂S₃) = 0.002695 mol Fe₂S₃ (unadjusted for percent yield).
Since the reaction has a percent yield of 65.0%, we adjust the moles of Fe₂S₃ for the actual yield:
0.002695 mol Fe₂S₃ / 0.65 = 0.004146 mol Fe₂S₃ (theoretical yield).
Using the balanced chemical equation 3 Li₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 LiCl(aq), we can determine the mole ratio of Fe₂S₃ to FeCl₃, which is 1:2. Therefore, we need 0.004146 mol Fe₂S₃ * 2 mol FeCl₃/mol Fe₂S₃ = 0.008292 mol FeCl₃.
Finally, we convert the moles of FeCl₃ to volume using its molarity (0.200 M):
0.008292 mol FeCl₃ / 0.200 mol/L = 0.04146 L, which equals 41.46 mL (rounded to three significant figures).
Therefore, 41.46 mL of 0.200 m FeCl₃ solution is required to react with an excess of Li₂S to theoretically produce 1.38 g of Fe₂S₃, considering a 65.0% yield.