(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 $Cu = \frac{1.27 \,g}{2.0 \,g} \times 100 = 63.5 \%$.