Question

2NH3 + CO2 -> CO(NH2)2 + H2O

Consider the reaction above, showing the production of the fertilizer urea, CO(NH2)2. What's the percent yield of urea if 800.0 grams of NH3
reacts with excess carbon dioxide to produce 1101 g?

Answer 1

To find the percent yield of urea, you need to compare the actual yield to the theoretical yield. The percent yield can be calculated using the formula (Actual yield / Theoretical yield) * 100%. In this case, the percent yield is 98.2%.

Step-by-step explanation:

To find the percent yield of urea, we need to compare the actual yield to the theoretical yield.

The balanced equation for the reaction is 2NH3 + CO2 -> CO(NH2)2 + H2O.

First, calculate the molar mass of NH3: 14.01 g/mol (N) + 3(1.008 g/mol) = 17.03 g/mol.

Next, calculate the moles of NH3 using the given mass: 800.0 g / 17.03 g/mol = 46.996 mol.

From the balanced equation, we see that 2 moles of NH3 produce 1 mole of urea (CO(NH2)2).

Thus, the theoretical yield of urea is 46.996 mol / 2 = 23.498 mol.

Finally, calculate the percent yield: (Actual yield / Theoretical yield) * 100%.

Assuming the actual yield is 1101 g, the percent yield would be (1101 g / (23.498 mol * 60.06 g/mol)) * 100% = 98.2%.