Question

Nitrogen gas, N2, is commonly used during samle preparation in chemical analysis to concentrate and reduce the volue of liquid samples. It can be prepared by the reaction between solid copper (III) oxide, CuO, and gaseous ammonia, NH3, at high temperatures. The other products of the reaction are solid copper, Cu, and water vapor. In an experiment, a reaction mixture containing 68.8 g CuO and 45.9 g NH3 and yields 4.25 g N2. Calculate the percent yield of the experiment.

Answer 1

39.7%

Step-by-step explanation:

The balanced chemical equation for the reaction is:

3CuO + 2NH3 → 3Cu + N2 + 3H2O

First, we need to calculate the theoretical yield of N2:

Molar mass of CuO = 79.55 g/mol

Molar mass of NH3 = 17.03 g/mol

Molar mass of N2 = 28.02 g/mol

Number of moles of CuO = 68.8 g / 79.55 g/mol = 0.865 mol

Number of moles of NH3 = 45.9 g / 17.03 g/mol = 2.695 mol

From the balanced equation, 3 moles of CuO react with 2 moles of NH3 to produce 1 mole of N2. Therefore, the number of moles of N2 produced should be:

(1 mol N2 / 3 mol CuO) x (0.865 mol CuO) x (2 mol NH3 / 3 mol CuO) = 0.383 mol N2

The theoretical yield of N2 is:

Theoretical yield = 0.383 mol x 28.02 g/mol = 10.73 g

The percent yield is:

Percent yield = (Actual yield / Theoretical yield) x 100%

Percent yield = (4.25 g / 10.73 g) x 100%

Percent yield = 39.7%

Therefore, the percent yield of the experiment is 39.7%.

Hope this helps!