From 50 g of nitrogen containing 5% impurities, 8 g of ammonia were obtained. Calculate the mass fraction

From 50 g of nitrogen containing 5% impurities, 8 g of ammonia were obtained. Calculate the mass fraction of the yield of ammonia from the theoretical yield of ammonia – more than 8 g.  

Let’s find the mass of pure nitrogen, without impurities.

100% – 5% = 95%.

50 g – 100%,

X g – 95%,

X = (95% × 50g): 100% = 47.5g.

Let’s find the amount of nitrogen substance.

n = m: M.

M (N2) = 28 g / mol.

n = 47.5 g: 28 g / mol = 1.696 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

3H2 + N2 = 2NH3

According to the reaction equation, there is 2 mol of ammonia per mole of hydrogen. The substances are in quantitative ratios of 1: 2.

The amount of ammonia substance will be 2 times more than the amount of nitrogen substance.

n (N2) = 2 n (NH3) = 1.696 × 2 = 3.39 mol.

Let’s find the mass of ammonia.

m = n × M.

M (NH3) = 17 g / mol.

m = 17 g / mol × 3.39 mol = 57.63 g.

57.63 g should be obtained theoretically (according to calculations).

According to the condition of the problem, it turned out 8 g.

Let’s find the output of ammonia from the theoretically possible.

57.63 g – 100%,

8 – x%,

X = (8 × 100): 57.63 = 13.88%.

Answer: 13.88%.