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Question: Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reactio...

Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction:
4Ag+8KCN+O2+2H2O4  K[Ag(CN)2]+4KOH4{\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 KCNKCN required to dissolve 100  g100\;{\text{g}} of pure Ag{\text{Ag}} is 120  g120\;{\text{g}}
(B) The amount of oxygen used in this process is 7.40  g7.40\;{\text{g}} (for 100gm100{\text{gm}} pure Ag{\text{Ag}} )
(C) The volume of oxygen used at S.T.P is 5.205.20 litres for the same process
(D) All of the above

Explanation

Solution

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
4Ag+8KCN+O2+2H2O4  K[Ag(CN)2]+4KOH4{\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
nAg4=nKCN8=nO21\dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{KCN}}}}{8} = \dfrac{{{n_{{O_2}}}}}{1}
Where
nn 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 Ag{\text{Ag}} is given as 100  g100\;{\text{g}}
The amount of KCNKCN required is given as 120  g120\;{\text{g}}
Now, let us check the validity of the option (A).
nAg4=nKCN8\dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{KCN}}}}{8}
Let the assumed mass of KCNKCN be mm
Then,
(100108)4=(m65)8\dfrac{{\left( {\dfrac{{100}}{{108}}} \right)}}{4} = \dfrac{{\left( {\dfrac{m}{{65}}} \right)}}{8}
Upon solving, we get
m=120.3g120gm = 120.3g \approx 120g
So, option (A.) is correct
Similarly,
For option (B.),
Let us assume the mass of O2{O_2} as mO2{m_{{O_2}}}
Then,
nAg4=nO21\dfrac{{{n_{Ag}}}}{4} = \dfrac{{{n_{{O_2}}}}}{1}
Upon substituting values, we get
(100108)4=(mO232)1\dfrac{{\left( {\dfrac{{100}}{{108}}} \right)}}{4} = \dfrac{{\left( {\dfrac{{{m_{{O_2}}}}}{{32}}} \right)}}{1}
On further solving, we get
mO2=7.40g{m_{{O_2}}} = 7.40g
So, option (B.) is also correct
Now, we will verify option (C.)
Volume of O2{O_2} at STP is given by
VO2(atSTP)=nO2×22.4{V_{{O_2}}}(at STP) = {n_{{O_2}}} \times 22.4
Where
nO2{n_{{O_2}}} is the number of moles of oxygen
Upon substituting the values, we get
VO2=7.40×22.4=5.25litre{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.