Question

2NH3(g) + 4H₂0 (1)→ 2NO2 + 7H2

If 85.0 grams of ammonia (NH3) are reacted with an excess of water, what would be the percent yield of nitrogen dioxide (NO2) if 215 grams are recovered. chart sample​

Answer 1

The percent yield of NO2 is 47.1%.

Step-by-step explanation:

To calculate the theoretical yield of NO2, we must first determine the limiting reactant. The balanced chemical equation tells us that 2 moles of NH3 react with 4 moles of H2O to produce 2 moles of NO2 and 7 moles of H2.

First, we need to calculate the number of moles of NH3 in 85.0 grams:

85.0 g NH3 x (1 mole NH3/17.03 g NH3) = 4.99 moles NH3

Since there is an excess of water, we assume that all of the NH3 will react to produce NO2. So, the theoretical yield of NO2 can be calculated as follows:

4.99 moles NH3 x (2 moles NO2/2 moles NH3) x (46.01 g NO2/1 mole NO2) = 455.9 g NO2

Next, we can calculate the percent yield:

Percent yield = (actual yield/theoretical yield) x 100%

Percent yield = (215 g NO2/455.9 g NO2) x 100% = 47.1%

Therefore, the percent yield of NO2 is 47.1%.