In a lab experiment, 17.5 g of iron II bromide react with 14.2 g of potassium sulfide.

Question

In a lab experiment, 17.5 g of iron II bromide react with 14.2 g of potassium sulfide. Upon completion, 4.3 g of iron II sulfide were obtained. Determine the limiting reactant, theoretical yield and percent yield.

Answer 1

FeBr₂ + K₂S ==> FeS + 2KBr ... This is the balanced equation you are dealing with.

First, determine which reactant, if any, is limiting:

moles FeBr₂ present = 17.5 g x 1 mole/215.7 g = 00811 moles

moles K₂S present = 14.2 g x 1 mole/110.3 g = 0.129 moles

Based on mole ratio of 1FeBr₂ : 1K₂S, the limiting reactant is FeBr₂

Theoretical yield of FeS = 0.0811 moles FeBr₂ x 1 mole FeS/mole FeBr₂ = 0.0811 moles

Theoretical mass yield of FeS = 0.0881 moles x 87.9 g/mole = 7.13 g = theoretical yield

Percent yield = 4.3 g/7.13 g (x100%) = 60% yield (to 2 significant figures; 60.3% to 3 sig. figs.)