Question

The weight of mixture containing \[HCl\] and \[{H_2}S{O_4}\] is $0.1g$ on treatment with an excess of an $AgN{O_3}$ solution, reacted with this acid mixture gives $0.1435\,g$ of $AgCl$ , weight percentage of \[{H_2}S{O_4}\] in mixture is:
A. $36.5$
B. $63.5$
C. $50$
D. $80$

Answer 1

Hint: For finding out the mass percent of sulphuric acid we have to firstly find out the mass of sulphuric acid and then by dividing it with the total mass of mixture we can easily calculate the mass percentage. Now for finding out the mass we have given another information that is mass of $AgCl$ which can be used for mass of it.

Complete step-by-step answer:The reaction of acid mixture with the $AgN{O_3}$ solution gives the product as $AgCl$ . We got only this product and not a sulphate product thus we can say it is less reactive, sulphuric acid is less reactive with silver nitrate. Let’s see their reaction firstly, so when acid mixture reacts with silver nitrate we get the reaction
$Acid\,mixture\,(HCl\, + \,{H_2}S{O_4})\, + \,AgN{O_3}\, \to \,AgCl\, + \,HN{O_3}$
We got silver chloride and nitric acid after the reaction also as mentioned in the question that the amount of silver chloride we get is $0.1435\,g$ .
$HCl + \,AgN{O_3}\, \to \,AgCl\, + \,HN{O_3}$
With the help of stoichiometry we can estimate the moles of \[HCl\] , $AgCl$ and $AgN{O_3}$ by the mass of given quantities. In the above reaction we see that one mole of hydrochloric acid reacts with one mole silver nitrate and gives one mole of silver chloride and one mole of nitric acid. As we see the number of moles are equal so we can easily find out the exact values of the number of moles of \[HCl\] from its given mass.
${n_{AgCl}} = \,{n_{HCl}} = \,{n_{AgN{O_3}}}$
${n_{HCl}} = \,\dfrac{{0.1435}}{{molar\,mass}}\, = \,\dfrac{{0.1435}}{{143.5}} = \,{10^{ - 3}}$
Weight of \[HCl\] is ${10^{ - 3}}\, \times \,36.5$ by the formula that moles is equals to the given mass divided by molar mass.
$Mass\,of\,HCl\, = \,0.0365\,g$
And we have given that mass of mixture is $0.1g$ then the mass of sulphuric acid will be $mass\,of\,{H_2}S{O_4}\, = \,0.1g\, - \,0.0365g\, = \,0.0635\,g$ . Mass percent of sulphuric acid can be defined as by dividing the mass of sulphuric acid with total mass and all last multiplying it with hundred.
$Mass\,\% \,of\,{H_2}S{O_4}\, = \,\dfrac{{Mass\,of\,{H_2}S{O_4}\,}}{{total\,mass\,}}\, \times 100$
 = $\,\dfrac{{0.0635\,}}{{0.1}}\, \times 100$
= $63.5\,\% $

Therefore option B is correct.

Note:As the number of moles are equal it means all reactants and products combine in same stoichiometry that is one mole hence we can write them equal, if the stoichiometry does not same then we cannot write the expression of moles equality. By that we get an estimation about the mass of hydrochloric acid.