
A quantity of $10{\text{g}}$ of a sample of silver which is contaminated with silver sulphide, gave $11.2{\text{mL}}$ of hydrogen sulphide at ${0^{\text{o}}}{\text{C}}$ and $1{\text{atm}}$, on treatment with excess of hydrochloric acid, The amount of silver sulphide in the sample is:
Given: mass of ${\text{Ag}} = 108$
A.$1.24{\text{g}}$
B.$124{\text{mg}}$
C.$5 \times {10^{ - 4}}{\text{mol}}$
D.$62{\text{mg}}$
Answer
579.3k+ views
Hint: 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,
${\text{Ag}} + {\text{A}}{{\text{g}}_{\text{2}}}{\text{S}} + {\text{HCl}} \to {{\text{H}}_{\text{2}}}{\text{S}}$
One mole of silver sulphide gives one mole of hydrogen sulphide.
We are given that the amount of hydrogen sulphide obtained is $11.2{\text{mL}}$.
We know that, at standard temperature and pressure, that is ${0^{\text{o}}}{\text{C}}$ and $1{\text{atm}}$, the volume occupied by one mole of a gas $ = 22.4{\text{L}} = 22400{\text{mL}}$
Or in other terms, one milliliter of gas at STP contains $\dfrac{1}{{22400}}{\text{moles}}$.
So, we can say that, $11.2{\text{mL}}$ of a gas at STP contains $11.2 \times \dfrac{1}{{22400}} = 5 \times {10^{ - 4}}{\text{moles}}$
Therefore, we have number of moles of hydrogen sulphide $ = 5 \times {10^{ - 4}}{\text{moles}}$
We can also conclude that, number of moles of silver sulphide $ = 5 \times {10^{ - 4}}{\text{moles}}$
Therefore, the weight of silver sulphide is the number of moles multiplied by the molar mass,
We get, ${\text{w}} = \left( {5 \times {{10}^{ - 4}}} \right) \times \left( {248} \right)$
$ \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.
Complete step by step answer:
For the given question, the reaction taking place is,
${\text{Ag}} + {\text{A}}{{\text{g}}_{\text{2}}}{\text{S}} + {\text{HCl}} \to {{\text{H}}_{\text{2}}}{\text{S}}$
One mole of silver sulphide gives one mole of hydrogen sulphide.
We are given that the amount of hydrogen sulphide obtained is $11.2{\text{mL}}$.
We know that, at standard temperature and pressure, that is ${0^{\text{o}}}{\text{C}}$ and $1{\text{atm}}$, the volume occupied by one mole of a gas $ = 22.4{\text{L}} = 22400{\text{mL}}$
Or in other terms, one milliliter of gas at STP contains $\dfrac{1}{{22400}}{\text{moles}}$.
So, we can say that, $11.2{\text{mL}}$ of a gas at STP contains $11.2 \times \dfrac{1}{{22400}} = 5 \times {10^{ - 4}}{\text{moles}}$
Therefore, we have number of moles of hydrogen sulphide $ = 5 \times {10^{ - 4}}{\text{moles}}$
We can also conclude that, number of moles of silver sulphide $ = 5 \times {10^{ - 4}}{\text{moles}}$
Therefore, the weight of silver sulphide is the number of moles multiplied by the molar mass,
We get, ${\text{w}} = \left( {5 \times {{10}^{ - 4}}} \right) \times \left( {248} \right)$
$ \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.
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