Hydrogen with a volume of 72 liters reacted with nitrogen. The ammonia yield is 85%.

Hydrogen with a volume of 72 liters reacted with nitrogen. The ammonia yield is 85%. Calculate the mass of the ammonia obtained.

Given:

VH2 = 72 l

vNH3 = 85%

72L x

N2 + 3 H2 = 2 NH3

1 mol 3 mol 2 mol

nH2: n NH3 = 3: 2

m NH3 -?

n = V / V mol. , where Vmol. = 22.4 liters the volume of one mole of any gas (molar volume).

nH2 = 72 / 22.4 = 3.2 mol,

Accordingly, n NH3 = 3.2 / 3 * 2 = 2.1 mol

m = M * n, where M is the molar mass of ammonia

MNH3 = 14 + 3 = 17 g / mol

Respectively

m NH3 = 2.1 * 17 = 36.3 g. mass of ammonia according to the reaction equation (theoretical yield).

v = (mn / mtot.) * 100%

Otsuda

m clean = v * mtot. / 100%

m NH3 = 85% * 36.3 / 100% = 30.9g.

Answer: 30.9 grams is the mass of the obtained ammonia.