For the reaction,#2x#+#3y#+#4z## rarr## 5w# Initially if 1 mole of #x#,3 mole of #y# and 4 mole of #z# is taken and 1.25 mole of #w# is obtained then what is the % yield of this reaction?

1 Answer
Nov 15, 2016

Answer:

The percent yield is 50 %.

Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation and moles of each reactant.

1. Gather all the information in one place with the experimental number of moles below the formulas.

#color(white)(mmmmmm)2x + 3y + 4z = 5w#
#"Amt/mol:"color(white)(ml)1color(white)(mm)3"color(white)(mm)4#
#"Divide by:"color(white)(ml)2color(white)(mm)3color(white)(mm)4#
#"Moles rxn:"color(white)(m)0.5color(white)(ml)1color(white)(mm)1#

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#x# is the limiting reactant because it gives the fewest moles of reaction.

3. Calculate the theoretical yield of #w#

#"Theoretical yield" = 1 color(red)(cancel(color(black)("mol"color(white)(l) x))) × ("5 mol " w)/(2 color(red)(cancel(color(black)("mol "x)))) = "2.5 mol"color(white)(l)w#

4. Calculate the percent yield

#"Percent yield" = "Actual yield"/"Theoretical yield" × 100 % = (1.25 color(red)(cancel(color(black)("mol"))))/(2.5 color(red)(cancel(color(black)("mol")))) × 100 % = 50 %#