Question

How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 1.38 g of Fe2S3 if the percent yield for the reaction is 65.0%? 3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq) A. 102 mL B. 25.5 mL C. 43.1 m D. 51.1 mL

Answer 1

To produce 1.38 g of Fe2S3 with a reaction per cent yield of 65.0%, we require 25.5 mL of 0.200 M FeCl3 solution.

Step-by-step explanation:

The student is tasked with determining the volume of 0.200 M FeCl3 solution required to react with an excess of Na2S to produce 1.38 g of Fe2S3 given a reaction per cent yield of 65%. Using the mole-mass calculation sequence, we first need to calculate the theoretical yield of Fe2S3.

To find the moles of Fe2S3, we use its molar mass:

(1.38 g Fe2S3) / (207.88 g/mol Fe2S3) = 0.00664 mol Fe2S3

Next, we adjust for the per cent yield:

0.00664 mol / 0.65 = 0.0102 mol Fe2S3 (theoretical yield)

Using the balanced equation 3 Na2S + 2 FeCl3 → Fe2S3 + 6 NaCl, we find the stoichiometric ratio of FeCl3 to Fe2S3 is 2:1. Therefore, we need 0.00510 mol FeCl3.

Finally, we calculate the required volume of FeCl3 solution:

(0.00510 mol FeCl3) / (0.200 mol/L) = 0.0255 L, which is 25.5 mL.

Thus, the answer is B. 25.5 mL.

Answer 2

To determine the amount of FeCl3 needed to react with an excess of Na2S and produce 1.38 g of Fe2S3 with a 65.0% yield, you can use stoichiometry and the concept of percent yield. First, calculate the moles of Fe2S3 produced, then use the balanced chemical equation to determine the moles of FeCl3 required. Finally, convert the moles of FeCl3 to milliliters using its molarity.

Step-by-step explanation:

To determine the amount of FeCl3 needed to react with an excess of Na2S, we need to use stoichiometry and the concept of percent yield. First, we calculate the moles of Fe2S3 produced using the molar mass of Fe2S3 and the given mass of Fe2S3. Then, using the balanced chemical equation, we can determine the moles of FeCl3 required. Finally, we convert the moles of FeCl3 to milliliters using the molarity of the FeCl3 solution.

The equation is balanced as:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

Given the percent yield of 65.0%, we calculate the moles of Fe2S3 produced:

Moles of Fe2S3 = (mass of Fe2S3 / molar mass of Fe2S3) × (1 / percent yield)

Using the stoichiometry of the balanced equation, we can determine the moles of FeCl3 required:

Moles of FeCl3 = (moles of Fe2S3 / 2) × (2 / 3)

Finally, we convert the moles of FeCl3 to milliliters using its molarity:

Milliliters of FeCl3 = (moles of FeCl3 / molarity of FeCl3) × 1000