Calculate the mass of(NH2)2CO that could theoretically be formed by this reaction.

Question

Sorry, I could not add a comment, Urea (NH₂)₂CO is prepared by reacting ammonia with carbon dioxide according to the following equation: 2NH₃(g) + CO₂(g) → (NH₂)₂CO(aq) + H₂O(l). In one experiment, 637.2 g of NH₃ is allowed to react with 1142 g of CO₂.

Answer 1

2NH3(g) + CO2(g) ==> (NH2)2CO(aq) + H2O(l).

Whenever given amounts of both reactants, you must first find the limiting reactant:

One way is to divide mols of each reactant by the corresponding coefficient and whichever value is less is the limiting reactant.

For NH₃: 637.2 g NH₃ x 1 mol NH₃ / 17 g = 37.48 mols NH₃ (÷2->18.74)

For CO₂: 1142 g CO₂ x 1 mol CO₂ / 44 g = 25.95 mols CO₂ (÷1->25.95)

Since 18.74 is less than 25.95, NH₃ is LIMITING

Now use moles of NH₃ (37.48 mols) to find theoretical yield of (NH₂)₂CO:

37.48 mols NH₃ x 1 mol (NH₂)₂CO / 2 mols NH₃ = 18.74 mols (NH₂)₂CO

molar mass (NH₂)₂CO = 60.06 g/mol

grams (NH₂)₂CO formed = 18.74 mols (NH₂)₂CO x 60.06 g / mol = 1126 g (NH₂)₂CO

Answer 2

First we calculate the number of moles on each reactant to find the limiting reagent.

n ( NH₃) = 632.7 ÷ 17

= 37.2 moles

n ( CO₂) = 1 142 ÷ 44

= 26 moles

This shows us that CO₂ was the limiting reagent in the reaction since it had fewer moles hence 26 moles reacted. We now come to the chemical equation to find the molar ratios. The ratio of CO₂ as to (NH₂)₂CO is 1 : 1. Therefore 26 moles of Urea were also formed.

Now we calculate the mass of Urea using the formula mass = n × Mr ( urea )

m (urea) = 26 × 60

= 1560 grams.