Question

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 142.1 kg of ammonia with 211.4 kg of carbon dioxide and obtains 171.4 kg of urea. Part A Determine the limiting reactant. Express your answer as a chemical formula.

Answer 1

Ammonia is the limiting reactant in the synthesis of urea from ammonia and carbon dioxide.

Step-by-step explanation:

The reaction between ammonia (NH3) and carbon dioxide (CO2) produces urea (CH4N2O) and water (H2O) according to the balanced equation: 2NH3(aq) + CO2(aq) → CH4N2O(aq) + H2O(l).

To determine the limiting reactant, we need to compare the amount of each reactant used to the amount of urea produced. From the given information, 142.1 kg of ammonia and 211.4 kg of carbon dioxide react to produce 171.4 kg of urea.

We can use the stoichiometry of the balanced equation to find the theoretical yield of urea from both reactants:

1. For ammonia: 1 mole of urea is produced from 2 moles of ammonia. The molar mass of ammonia is 17.03 g/mol, so the number of moles of ammonia in 142.1 kg is (142.1 kg) / (17.03 g/mol) = 8358.4 mol. Therefore, the theoretical yield of urea from ammonia is (8358.4 mol) / 2 = 4179.2 mol or 4179.2 mol × 60.06 g/mol = 250783.4 g.
2. For carbon dioxide: 1 mole of urea is produced from 1 mole of carbon dioxide. The molar mass of carbon dioxide is 44.01 g/mol, so the number of moles of carbon dioxide in 211.4 kg is (211.4 kg) / (44.01 g/mol) = 4801.6 mol. Therefore, the theoretical yield of urea from carbon dioxide is 4801.6 mol × 60.06 g/mol = 288255.4 g.

Since the actual yield of urea is 171.4 kg, which is less than both the theoretical yields from ammonia and carbon dioxide, the limiting reactant is ammonia (NH3).

Therefore, the limiting reactant is ammonia (NH3).

Answer 2

The limiting reactant is Carbon dioxide,
`CO_(2)`

Step-by-step explanation:

The balanced reaction equation is:

`2NH_(3) + CO_(2)` →
`CH_(4) N_(2) O + H_(2) O`

The mole ratio of ammonia to carbon dioxide is 2:1

142100/17g = 8358.8 mol of NH3

211400/44g = 4, 804.5 mole of CO2

Now:

4,804.5 mol of CO2 ×
`(2 mol NH_(3) )/(1 molCO_(2) )` = 9,609 mol of NH3 present

8,358.8 mole of NH3 ×
`(1 moleCO_(2) )/(2moles NH_(3) )` = 4,179.4 mol of CO2 present

NH3 needs 8, 358.8 moles but had 9, 609 moles⇒ excess reactant

CO2 needs 4, 804,5 mol but had 4, 179.4 moles⇒ limiting reactant (used up completely)

The limiting reactant is carbon dioxide, CO2.