Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
