Question

A common method to produce ammonium nitrate, an important agricultural fertilizer, is to combine ammonium sulfate, (NH4)2SO4, with sodium nitrate, NaNO3, as shown in the following balanced equation. (NH4)2SO4(s) + 2 NaNO3(s) → Na2SO4(s) + 2 NH4NO3(s) If 681.4 g (NH4)2SO4 is reacted with excess NaNO3 and produces 748.6 g NH4NO3, what is the percent yield?

Answer 1

Molar mass of (NH4)2SO4,

MM = 2*MM(N) + 8*MM(H) + 1*MM(S) + 4*MM(O)

= 2*14.01 + 8*1.008 + 1*32.07 + 4*16.0

= 132.154 g/mol

mass((NH4)2SO4)= 681.4 g

number of mol of (NH4)2SO4,

n = mass of (NH4)2SO4/molar mass of (NH4)2SO4

=(681.4 g)/(132.154 g/mol)

= 5.156 mol

Balanced chemical equation is:

(NH4)2SO4 + 2 NaNO3 ---> 2 NH4NO3 + Na2SO4

Molar mass of NH4NO3,

MM = 2*MM(N) + 4*MM(H) + 3*MM(O)

= 2*14.01 + 4*1.008 + 3*16.0

= 80.052 g/mol

According to balanced equation

mol of NH4NO3 formed = (2/1)* moles of (NH4)2SO4

= (2/1)*5.1561

= 10.31 mol

mass of NH4NO3 = number of mol * molar mass

= 10.31*80.05

= 825.5 g

% yield = actual mass*100/theoretical mass

= 748.6*100/825.5

= 90.7 %

Answer: 90.7 %