Question

calculate the percent yield of this reaction if 623g of ammonia produces 2341g of ammonium bromide

Answer 1

3Br₂ + 8NH₃ → N₂ + 6NH₄Br

Molar mass of ammonia is 17.03g/mol

Molar mass of ammonium bromide is 97.94g/mol

Mole ratio of ammonia to ammonium bromide is 8:6

$136gNH_3$ gives $587.64gNH_4Br$

so$623g NH_3$ gives

$587.64×\frac{623}{136} =2691.9g NH_4Br$

but it actually produces $2341g NH_4Br$

Hence the percent yield will be

$\frac{2341}{2691.9} ×100 = 86.96\%$