Copper in an alloy is estimated by dissolving in conc. nitric acid. In this process,copper is converted to cupric nitrate with the evolution of nitric oxide $(NO)$. The mixture when treated with potassium iodide forms cupric iodide,which is unstable and decomposes to cuprous iodide and iodine. The amount of copper in the alloy is estimated by titrating the liberated iodine with sodium thiosulphate. The reactions are:
$a \,Cu + b \,HNO_3 \rightarrow c \,Cu(NO_3)_2 + d \,NO + e \,H_2O$
$f \,CuI_2 \rightarrow g \,Cu_2I_2 + h \,I_2$
$i \,Na_2S_2O_3 + j \,I_2 \rightarrow k \,Na_2S_4O_6 + l \,NaI$
(Fill in the blanks)
$(a)$ The coefficients are: $a=\ldots, b=\ldots, c=\ldots, d=\ldots$ and $e=\ldots$.
$(b)$ The coefficients are: $f=\ldots, g=\ldots$ and $h=\ldots$.
$(c)$ The coefficients are: $i=\ldots, j=\ldots, k=\ldots$ and $l=\ldots$.
$(d)$ If $2.54 \,g$ of $I_2$ is evolved from a $2.0 \,g$ sample of the alloy,what is the percentage of copper in the alloy? (Atomic weights of iodine and copper are $127$ and $63.5$,respectively).
$a \,Cu + b \,HNO_3 \rightarrow c \,Cu(NO_3)_2 + d \,NO + e \,H_2O$
$f \,CuI_2 \rightarrow g \,Cu_2I_2 + h \,I_2$
$i \,Na_2S_2O_3 + j \,I_2 \rightarrow k \,Na_2S_4O_6 + l \,NaI$
(Fill in the blanks)
$(a)$ The coefficients are: $a=\ldots, b=\ldots, c=\ldots, d=\ldots$ and $e=\ldots$.
$(b)$ The coefficients are: $f=\ldots, g=\ldots$ and $h=\ldots$.
$(c)$ The coefficients are: $i=\ldots, j=\ldots, k=\ldots$ and $l=\ldots$.
$(d)$ If $2.54 \,g$ of $I_2$ is evolved from a $2.0 \,g$ sample of the alloy,what is the percentage of copper in the alloy? (Atomic weights of iodine and copper are $127$ and $63.5$,respectively).
(D) The balanced chemical equations are:
$(i)$ $3 \,Cu + 8 \,HNO_3 \rightarrow 3 \,Cu(NO_3)_2 + 2 \,NO + 4 \,H_2O$
$(ii)$ $2 \,CuI_2 \rightarrow Cu_2I_2 + I_2$
$(iii)$ $2 \,Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 \,NaI$
$(a)$ The coefficients are: $a=3, b=8, c=3, d=2, e=4$.
$(b)$ The coefficients are: $f=2, g=1, h=1$.
$(c)$ The coefficients are: $i=2, j=1, k=1, l=2$.
$(d)$ Moles of $I_2 = \frac{2.54 \,g}{254 \,g/mol} = 0.01 \,mol$.
From equation $(ii)$,$1 \,mol$ of $I_2$ is produced from $2 \,mol$ of $CuI_2$,which corresponds to $2 \,mol$ of $Cu$.
Therefore,moles of $Cu = 2 \times 0.01 \,mol = 0.02 \,mol$.
Mass of $Cu = 0.02 \,mol \times 63.5 \,g/mol = 1.27 \,g$.
Percentage of $Cu = \frac{1.27 \,g}{2.0 \,g} \times 100 = 63.5 \%$.
$(i)$ $3 \,Cu + 8 \,HNO_3 \rightarrow 3 \,Cu(NO_3)_2 + 2 \,NO + 4 \,H_2O$
$(ii)$ $2 \,CuI_2 \rightarrow Cu_2I_2 + I_2$
$(iii)$ $2 \,Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 \,NaI$
$(a)$ The coefficients are: $a=3, b=8, c=3, d=2, e=4$.
$(b)$ The coefficients are: $f=2, g=1, h=1$.
$(c)$ The coefficients are: $i=2, j=1, k=1, l=2$.
$(d)$ Moles of $I_2 = \frac{2.54 \,g}{254 \,g/mol} = 0.01 \,mol$.
From equation $(ii)$,$1 \,mol$ of $I_2$ is produced from $2 \,mol$ of $CuI_2$,which corresponds to $2 \,mol$ of $Cu$.
Therefore,moles of $Cu = 2 \times 0.01 \,mol = 0.02 \,mol$.
Mass of $Cu = 0.02 \,mol \times 63.5 \,g/mol = 1.27 \,g$.
Percentage of $Cu = \frac{1.27 \,g}{2.0 \,g} \times 100 = 63.5 \%$.
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