From 56 g of nitrogen, 48 g of ammonia was obtained. What is the yield of ammonia in% of theoretical?

Given:

m (N2) = 56 g

m (NH3) = 48 g

To find:

w% (NH3 practical) -?

Decision:

1) We solve the problem using the reaction equation compiled according to the condition:

N2 + 3H2 = 2NH3;

2) Find the amount of nitrogen contained in 56 grams of gas:

n (N2) = 56 g: 28 g / mol = 2 mol;

3) We compose logical equality:

if 1 mol of N2 gives 2 mol of NH3,

then 2 mol of N2 will give x mol of NH3,

then x = 4 mol.

4) Find the theoretically possible yield of ammonia:

m (NH3) = n * M = 4 mol * 17 g / mol = 68 g;

5) Find the ratio of theoretical and practical product yield:

w% (NH3 practical) = 48 g: 68 g * 100% = 70.6%.

Answer: w% (NH3 practical) = 70.6%.