Answer to Question #201989 in General Chemistry for James

Question

50.0g of ammonia reacted with 50.0g of carbon dioxide and 60.0g of urea was obtained.

Determine the percent yield for this reaction. Give the formula for the reaction and state 

the limiting reagent.

Answer 1

Reaction of ammonia with carbon dioxide 

2NH_3(g) + CO_2(g)$\implies$ NH₂CONH_2(s)+ H₂O_(l)

Mass of NH₃= 50.0g 

Mass of CO₂ = 50.0g 

Molar mass of NH₃= 17g/mol

Moles of NH₃ = 50.0g÷17gmol^-1

= 2.941 mol

Molar mass of CO₂ = 44g/mol

Moles of CO₂= 50.0g÷44gmol^-1 = 2.941 mol

Molar mass of CO₂ = 44g/mol

Moles of CO₂= 50.0g÷44gmol^-1 = 1.136mol

By reaction stochiometry : 

(1mol CO₂$\div$ 2mol NH₃) × 2.941mol NH₃ = 1.471 mol CO₂

But in reaction mixture only 1.136 mol CO₂ is present. So, CO₂will completely react during reaction.

Limiting reagent = CO₂

By reaction stochiometry: 

(1mol NH₂CONH₂$\div$ 1mol CO₂) ×1.136mol CO₂ = 1.136 mol NH₂CONH₂

1.136mol × 60g/mol urea = 68.2 g 

So, theortical yield of urea = 68.2g

Actual yield of urea = 60.0 g 

Percent yield = $\frac{60.0g}{68.2g}$ ×100 = 87.9% 

So, percent yield for the reaction = 87.9%