Solveeit Logo
Login

Question

Question: A quantity of \(10{\text{g}}\) of a sample of silver which is contaminated with silver sulphide, gav...

A quantity of 10g10{\text{g}} of a sample of silver which is contaminated with silver sulphide, gave 11.2mL11.2{\text{mL}} of hydrogen sulphide at 0oC{0^{\text{o}}}{\text{C}} and 1atm1{\text{atm}}, on treatment with excess of hydrochloric acid, The amount of silver sulphide in the sample is:
Given: mass of Ag=108{\text{Ag}} = 108
A.1.24g1.24{\text{g}}
B.124mg124{\text{mg}}
C.5×10−4mol5 \times {10^{ - 4}}{\text{mol}}
D.62mg62{\text{mg}}

Explanation

Solution

To solve this question, you must recall basic stoichiometric fundamentals. Stoichiometry is based on the law of conservation of mass which suggests that the total mass of reactants is equal to the total mass of products. Thus, if the amount of separate reactants are known to us, then the amount of products can be determined.

Complete step by step answer:
For the given question, the reaction taking place is,
Ag+Ag2S+HCl→H2S{\text{Ag}} + {\text{A}}{{\text{g}}_{\text{2}}}{\text{S}} + {\text{HCl}} \to {{\text{H}}_{\text{2}}}{\text{S}}
One mole of silver sulphide gives one mole of hydrogen sulphide.
We are given that the amount of hydrogen sulphide obtained is 11.2mL11.2{\text{mL}}.
We know that, at standard temperature and pressure, that is 0oC{0^{\text{o}}}{\text{C}} and 1atm1{\text{atm}}, the volume occupied by one mole of a gas =22.4L=22400mL = 22.4{\text{L}} = 22400{\text{mL}}
Or in other terms, one milliliter of gas at STP contains 122400moles\dfrac{1}{{22400}}{\text{moles}}.
So, we can say that, 11.2mL11.2{\text{mL}} of a gas at STP contains 11.2×122400=5×10−4moles11.2 \times \dfrac{1}{{22400}} = 5 \times {10^{ - 4}}{\text{moles}}
Therefore, we have number of moles of hydrogen sulphide =5×10−4moles = 5 \times {10^{ - 4}}{\text{moles}}
We can also conclude that, number of moles of silver sulphide =5×10−4moles = 5 \times {10^{ - 4}}{\text{moles}}
Therefore, the weight of silver sulphide is the number of moles multiplied by the molar mass,
We get, w=(5×10−4)×(248){\text{w}} = \left( {5 \times {{10}^{ - 4}}} \right) \times \left( {248} \right)
⇒w=0.124gm=124mg\Rightarrow {\text{w}} = 0.124{\text{gm}} = 124{\text{mg}}

Therefore, the correct option is B.

Note:
In general, chemicals combine in definite ratios in the chemical reactions. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, thus the amount of each element must be the same throughout the entire reaction. For instance, the number of atoms of a given element X on the reactant side must be equal to the number of atoms of that element on the product side, irrespective of whether or not all of those atoms are involved in a reaction.