How many mm of 0.200 M FeCl3 are needed to react with an excess of K2S to produce 2.75 g of Fe2S3 if the percent yield for the reaction is 65.0%?

130 mL

Explanation:

2FeCl3(aq) + 3K2S(aq) > Fe2S3(s) + 6KCl(aq)

Percent yield = actual yield/theoretical yield × 100

actual yield = 2.75 g

Theoretical yield = x

65= 2.75/x × 100

x= 2.75 × 100/65

x= 4.2 g

Number of moles of FeCl3 = mass/molar mass = 4.2 g/ 162.2 g/mol = 0.026 moles

Then;

n= CV

n= number of moles

C= concentration

V= volume

V= n/C

V= 0.026 moles/ 0.200 M

V= 0.13 L or 130 mL