Precent Yield with Limited reactant

Question

For the following reaction, 6.15 grams of water are mixed with 11.3 grams of carbon monoxide. What is the theoretical yield of carbon dioxide? CO + H2O -> CO2 + H2

 

Theoretical yield of CO2

LIMITED REAGENT

PRECENT YIELD?

Actual yield 12.8 grams is CO2.

Answer 1

CO + H₂O -> CO₂ + H₂

Determine limiting reactant:

moles CO =11.3 g CO x 1 mole CO/28 g = 0.404 moles CO

moles H₂O = 6.15 g x 1 mole H₂O/18 g = 0.342 moles H₂O

Since they are present in the balanced equation in a 1:1 mole ratio, H₂O is LIMITING.

 

Determine theoretical yield:

This will be determined by the moles of water, as it is limiting.

0.342 moles H₂O x 1 mole CO₂/mole H₂O = 0.342 moles CO₂ theoretical yield

0.342 moles CO₂ x 44 g/mole = 15.0 g CO₂ = theoretical yield

 

Determine percent yield:

This is actual yield/theoretical yield (x100)

12.8 g/15.0 g (x100) = 85.3% = percent yield