Question

: How many grams of ammonium sulfate can be produced if $60\,mol$ of sulfuric acid react with an excess of ammonia?
$2N{H_3}\,(aq)\, + \,{H_2}S{O_4}(aq)\, \to \,{(N{H_4})_2}S{O_4}\,(aq)$

Answer 1

The above problem is for the molar ratio, here $2\,mol$ of ammonia reacts with $1\,mol$ of sulfuric acid thus produced $1\,mol$ of ammonium sulfate. We need to understand the conversion of moles into mass as we want the number of grams of ammonium sulfate. Find out the moles formed according to the reaction and then convert them into grams.

Complete step-by-step answer:
According to the reaction, $2N{H_3}\,(aq)\, + \,{H_2}S{O_4}(aq)\, \to \,{(N{H_4})_2}S{O_4}\,(aq)$ here $2\,mol$ of ammonia reacts with $1\,mol$ of sulfuric acid thus produced $1\,mol$ of ammonium sulfate. So we can say by using mathematical equation that,
$1\,mol\,{H_2}S{O_4}\,produced\, = \,1\,mole\,of\,{(N{H_4})_2}S{O_4}\,$
$60\,mol\,{H_2}S{O_4}\,produced\, = \,60\,\,mole\,of\,{(N{H_4})_2}S{O_4}\,$
There is a stoichiometry of $1:1$ between ${H_2}S{O_4}$ and ${(N{H_4})_2}S{O_4}$ thus we can easily get to know about the number of moles of ammonium sulphate which will be formed after the reaction.
As one mole of sulfuric acid produced one mole of ammonium sulphate according to the stoichiometry thus $60\,mol$ of sulfuric acid produces $60\,mol$ of ammonium sulphate. Now the question arises how we will convert it in grams. So firstly find out the molar mass of ammonium sulphate.
$Molar\,mass\,of{(N{H_4})_2}S{O_4}\, = \,2 \times (14 + 4) + 32.07 + 4 \times (16)$
$= 132.17\,g\,mo{l^{ - 1}}$
It means that one mole of ${(N{H_4})_2}S{O_4}\,$ contains $132.17\,g$ of ${(N{H_4})_2}S{O_4}\,$ thus for $60\,mol$ of ${(N{H_4})_2}S{O_4}\,$ we have grams as,
$60\,mol\,of\,{(N{H_4})_2}S{O_4}\, = \,60 \times 132.17\,grams\,of{(N{H_4})_2}S{O_4}\,$
$= 7962\,grams\,$

Note: In these types of questions we can find out the number of particles also by multiplying the moles with Avogadro's number. So, the conversion is very important. We should know how to convert moles into mass and moles into a number of particles. As above the stoichiometry is of $1:1$ between ${H_2}S{O_4}$ and ${(N{H_4})_2}S{O_4}$ thus we easily found the number of moles but in certain cases the ratio is different.