How many moles of lead (II) chloride will be formed from a reaction between $6.5 \mathrm{~g}$ of $\mathrm{PbO}$ and $3.2 \mathrm{~g}$ of $\mathrm{HCl}$ ?

Key Idea : The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles $=\frac{\text { weight }}{\text { molecular weight }}$ $\begin{array}{ccc} \mathrm{PbO} & 2 \mathrm{HCl} \longrightarrow & \mathrm{PbCl}_2+\mathrm{H}_2 \mathrm{O} \\ 207.2+16 & 2(35.5+1) & 207.2+71 \\ =223.2 & =73 & =278.2 \end{array}$ Here, 1 mole of $\mathrm{PbO}$ reacts with 2 moles of $\mathrm{HCl}$, thus $\mathrm{PbO}$ is the limiting reagent $\because 223.2 \mathrm{~g}$ PbO gives $\mathrm{PbCl}_2=278.2 \mathrm{~g}$ $\therefore 6.5 \mathrm{~g} \mathrm{PbO}$ will give $\mathrm{PbCl}_2$ $\begin{aligned} & =\frac{278.2}{223.2} \times 6.5 \mathrm{~g} \\ & =\frac{278.2 \times 6.5}{223.2 \times 278.2} \mathrm{~mol} \\ & =0.029 \mathrm{~mol} \end{aligned}$