Chemistry - Mole Concept Question with Solution | TestHub

$$\underset{2249}{\mathrm{PbO}}+\underset{73\mathrm{g}}{2\mathrm{HCl}}⟶{\mathrm{PbCl}}_2+{\mathrm{H}}_2\mathrm{O}$$ Here, $\mathrm{HCl}$ is in excess, hence, $\mathrm{PbO}$ is the limiting reactant. $$\begin{array}{rl}& \because 224\mathrm{g}\mathrm{PbO}\equiv 1\text{ mole }{\mathrm{PbCl}}_2\\ & \therefore 6.5\mathrm{g}\mathrm{PbO}=\frac{6.5}{224}\text{ mole }{\mathrm{PbCl}}_2\\ & =0.029\mathrm{mole}\\ & \end{array}$$