Question

Equal mass of $Zn$ metal and iodine are mixed together and the iodine is completely converted to $Zn{I}_2$. What fraction of mass of the original $Zn$ remains unreacted?
(Write numerical value upto $2$ digits only after decimal point as for $0.19$, write $19$)

Answer 1

The molar masses of Zn and iodine are 65.39 g/mol and 253.8 g/mol respectively.
Let us assume that 253.8 g of zinc are mixed with 253.8 g of iodine.
The number of moles of Zn $=\frac{253.8}{65.39}=3.88$.
The number of moles of iodine $=\frac{253.8}{253.8}=1$
The balanced chemical equation is $Zn+I_2\rightleftharpoons Zn{I}_2$
Thus, 1 mole of Zn reacts with 1 mole of iodine.
The number of moles of unreacted Zn are $3.88-1=2.88$
The mass of unreacted zinc is $65.39\times 2.88=188.41$g.
The percentage of zinc unreactred is $\frac{188.41}{253.8}\times 100=74$ %