# Answer to Question #201989 in General Chemistry for James

Question #201989

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.

1
2021-06-02T06:31:58-0400

Reaction of ammonia with carbon dioxide

2NH3(g) + CO2(g)"\\implies" NH2CONH2(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%

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!