Question

The reaction shown below produced 1.38 g Fe2S3, corresponding to a percent yield for the reaction of 65%. An excess of Na2S was used in the reaction. How many milliliters of 0.200 M FeCl3 was used in the reaction? 3Na2S (aq) + 2FeCl3 (aq)-->Fe2S3 (s) + 6NaCl (aq)

A) 12.5 mL
B) 25.0 mL
C) 37.5 mL
D) 50.0 mL

Answer 1

To find the volume of 0.200 M FeCl3 solution used in the reaction, we need to calculate the moles of FeCl3 and use its molarity. The answer is B) 25.0 mL.

Step-by-step explanation:

In order to find the amount of FeCl3 used in the reaction, we need to use the concept of stoichiometry and percent yield.

Given that the percent yield of the reaction is 65% and that 1.38 g of Fe2S3 was produced, we can calculate the theoretical yield of Fe2S3 by dividing the actual yield by the percent yield:

Theoretical yield = actual yield / percent yield = 1.38 g / 0.65 = 2.12 g

Since Fe2S3 and FeCl3 have a 1:2 molar ratio according to the balanced equation, the moles of FeCl3 used can be calculated using the mole-mass calculation:

Moles of FeCl3 = mass of Fe2S3 / molar mass of Fe2S3 * (moles of FeCl3 / moles of Fe2S3)

Assuming the molar mass of Fe2S3 is 207.9 g/mol and the molar ratio is 2:1, we get:

Moles of FeCl3 = 2.12 g / 207.9 g/mol * (1 mol / 2 mol) = 0.051 mol

Finally, we can calculate the volume of the 0.200 M FeCl3 solution using the concentration of the solution:

Volume of FeCl3 = moles of FeCl3 / molarity of FeCl3 = 0.051 mol / 0.200 M = 0.255 L = 255 mL

Therefore, the answer is option B) 25.0 mL.