b
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

\(=\frac{\text { weight }}{\text { molecular weight }}\)

\(\underset{207.2+16=223.2}{\mathrm{PbO}}+\underset{2(35.5+1)=73}{ \mathrm{HCl}} \rightarrow\)\(\underset{207.2+71=278.2}{\mathrm{PbCl}_{2}}+\mathrm{H}_{2} \mathrm{O}\)

Here, \(1 \;mole\) of \(PbO\) reacts with \(2\; moles\) of \(HCl\), thus \(PbO\) is the limiting reagent

\(\because 223.2 \;\mathrm{g\;PbO}\) gives \(\mathrm{PbCl}_{2}=278.2 \mathrm{g}\)

\(\therefore 6.5 \mathrm{g} \;PbO\) will give \(\mathrm{PbCl}_{2}\)

\(=\frac{278.2}{232.2} \times 6.5 \;\mathrm{g}\)

\(=\frac{278.2 \times 6.5}{223.2 \times 278.2}\; \mathrm{mol}\)