Question

Calculate the mass of copper required to react 63 g of dilute nitric acid(Cu=64, H=1, N=12and O=16)

Answer 1

Step-1 Calculate the molar mass of Copper(Cu) and Nitric acid $\left({\mathrm{HNO}}_3\right)$

According to the reaction $3\mathrm{Cu}\left(\mathrm{s}\right)+8{\mathrm{HNO}}_3\left(\mathrm{aq}\right)\to 3\mathrm{Cu}{({\mathrm{NO}}_3)}_2\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2\mathrm{NO}\left(\mathrm{g}\right)$

$3$ moles of Copper (Cu) react with $8$ moles of Nitric acid $\left({\mathrm{HNO}}_3\right)$

Molar mass of $Cu=64g$

The molar mass of ${\mathrm{HNO}}_3=$Atomic mass of $H$ + Atomic mass of$N$+ $3\times$ Atomic mass of $\mathrm{O}$

$=1+14\times \left(3\times 16\right)=63g$

Step-2 Calculate the Overall mass of $Cu$used in the reaction

Total $Cu$used in the reaction= $3\times 64=192\mathrm{g}$

Total ${\mathrm{HNO}}_3$ used in the reaction= $=8\times 63=504\mathrm{g}$

If , $504\mathrm{g}$ of ${\mathrm{HNO}}_3$ requires $=192\mathrm{g}\mathrm{Cu}$ (According to the equation)

Then, $63\mathrm{g}$ of ${\mathrm{HNO}}_3$ requires $=\left(\frac{192\times 63}{504}\right)\mathrm{g}=24\mathrm{g}\mathrm{of}\mathrm{Cu}$

Hence, the mass of $Cu$required to react 63g of dilute Nitric acid $\mathrm{is}24\mathrm{g}$