Question

Urea, which has the chemical formula (NH2)2CO, is a fertilizer that can be prepared by reacting ammonia (NH3) with carbon dioxide (CO2). Given the following chemical equation, what is the theoretical yield of urea (in grams) if 4.02 mol ammonia is the limiting reactant? 2 NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l).

Answer 1

so 2 mole NH3 gives 1 mol urea
then 4.02 moles gives 2.01 moles of urea !!

so in gram = no of moles × molecular formula mass
= 2.01×60.06
= 120.72 grams ...your answer !!

Answer 2

The theoretical yield of urea is 121 grams.

Explanation:

Given:

The reaction between ammonia gas and carbon dioxide gas to prepare urea.

4.02 moles of ammonia gas is a limiting reactant.

To find:

The theoretical yield of urea.

Solution

$2 NH_3(g) + CO_2(g) \rightarrow (NH_2)_2CO(aq) + H_2O(l)$

The moles of ammonia gas = 4.02 mol

The ammonia gas is a limiting reactant so that means the theoretical yield of urea produced will depend upon moles of ammonia gas.

According to the reaction, 2 moles of ammonia gives 1 mole of urea, then 4.02 moles of ammonia will give:

$=\frac{1}{2}\times 4.02 mol=2.01 \text{mol of urea}$

The mass of 2.01 moles urea:

$=2.01 mol\times 60.06 g/mol=120.72 g\approx 121 g$

The theoretical yield of urea is 121 grams.

Learn more about limiting reactants here:

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