Question

How many milliliters of 0.250 M FeCl3 are needed to react with an excess of Na2S to produce 3.50 g of Fe2S3 if the percent yield for the reaction is 70.0%?

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

Answer 1

Actual yield = 3.50 g
Lets calculate theoretical yield
% yield = actual yield * 100 / theoretical yield
70.0 = 3.50 * 100 / theoretical yield
theoretical yield = 5.00 g

Molar mass of Fe2S3,
MM = 2*MM(Fe) + 3*MM(S)
= 2*55.85 + 3*32.07
= 207.91 g/mol


mass(Fe2S3)= 5.0 g

use:
number of mol of Fe2S3,
n = mass of Fe2S3/molar mass of Fe2S3
=(5 g)/(2.079*10^2 g/mol)
= 2.405*10^-2 mol
According to balanced equation
mol of FeCl3 reacted = (2/1)* moles of Fe2S3
= (2/1)*2.405*10^-2
= 4.81*10^-2 mol
This is number of moles of FeCl3

use:
M = number of mol / volume in L
0.25 = 4.81*10^-2/ volume in L
volume = 0.192 L
volume = 192 mL
Answer: 192 mL