Question

Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction:
 $ 4{\text{Ag}} + 8{\text{KCN}} + {{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}} \to 4\;{\text{K}}\left[ {{\text{Ag}}{{({\text{CN}})}_2}} \right] + 4{\text{KOH}} $
Which of the following is true regarding the above reaction?
(A) The amount of $ KCN $ required to dissolve $ 100\;{\text{g}} $ of pure $ {\text{Ag}} $ is $ 120\;{\text{g}} $
(B) The amount of oxygen used in this process is $ 7.40\;{\text{g}} $ (for $ 100{\text{gm}} $ pure $ {\text{Ag}} $ )
(C) The volume of oxygen used at S.T.P is $ 5.20 $ litres for the same process
(D) All of the above

Answer 1

Hint: The number of molecules that take part in the reaction is the stoichiometric coefficient or stoichiometric number. You will notice that there are an equal number of elements on both sides of the equation when you look at any balanced reaction. Basically, the stoichiometric coefficient is the number that is present before atoms, molecules, or ions. We can use these to also find the mass of a product formed when mass of reactants are given.

Complete Step-by-Step solution
Let us first write the equation that is provided in the question
 $ 4{\text{Ag}} + 8{\text{KCN}} + {{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}} \to 4\;{\text{K}}\left[ {{\text{Ag}}{{({\text{CN}})}_2}} \right] + 4{\text{KOH}} $
Now, we can say that
 $ \dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{KCN}}}}{8} = \dfrac{{{n_{{O_2}}}}}{1} $
Where
 $ n $ is the number of moles
And the numbers are the stoichiometric coefficients of the reactants of the chemical equation
Now, let us take a look at each of the options provided
The weight of $ {\text{Ag}} $ is given as $ 100\;{\text{g}} $
The amount of $ KCN $ required is given as $ 120\;{\text{g}} $
Now, let us check the validity of the option (A).
 $ \dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{KCN}}}}{8} $
Let the assumed mass of $ KCN $ be $ m $
Then,
 $ \dfrac{{\left( {\dfrac{{100}}{{108}}} \right)}}{4} = \dfrac{{\left( {\dfrac{m}{{65}}} \right)}}{8} $
Upon solving, we get
 $ m = 120.3g \approx 120g $
So, option (A.) is correct
Similarly,
For option (B.),
Let us assume the mass of $ {O_2} $ as $ {m_{{O_2}}} $
Then,
 $ \dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{{O_2}}}}}{1} $
Upon substituting values, we get
 $ \dfrac{{\left( {\dfrac{{100}}{{108}}} \right)}}{4} = \dfrac{{\left( {\dfrac{{{m_{{O_2}}}}}{{32}}} \right)}}{1} $
On further solving, we get
 $ {m_{{O_2}}} = 7.40g $
So, option (B.) is also correct
Now, we will verify option (C.)
Volume of $ {O_2} $ at STP is given by
 $ {V_{{O_2}}}(at STP) = {n_{{O_2}}} \times 22.4 $
Where
 $ {n_{{O_2}}} $ is the number of moles of oxygen
Upon substituting the values, we get
 $ {V_{{O_2}}} = 7.40 \times 22.4 = 5.25 litre $
So, option (C.) is also correct
Hence, the correct option is (D).

Note
Fractions as well as entire numbers can be stoichiometric coefficients. The coefficients help us, in essence, to determine the mole ratio between reactants and products.