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

The balanced chemical equation is: $\text{PbO} + 2\text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O}$

Molar mass of $\text{PbO} = 207.2 + 16 = 223.2\text{ g/mol}$. Moles of $\text{PbO} = \frac{6.5\text{ g}}{223.2\text{ g/mol}} \approx 0.0291\text{ mol}$.

Molar mass of $\text{HCl} = 1 + 35.5 = 36.5\text{ g/mol}$. Moles of $\text{HCl} = \frac{3.2\text{ g}}{36.5\text{ g/mol}} \approx 0.0876\text{ mol}$.

From the balanced equation, $1\text{ mole}$ of $\text{PbO}$ reacts with $2\text{ moles}$ of $\text{HCl}$. So, $0.0291\text{ moles}$ of $\text{PbO}$ will require $0.0291 \times 2 = 0.0582\text{ moles}$ of $\text{HCl}$. Since $0.0876\text{ moles}$ of $\text{HCl}$ are available, which is more than the required amount ($0.0582\text{ moles}$), $\text{PbO}$ is the limiting reagent and $\text{HCl}$ is in excess.

The amount of product formed depends completely on the limiting reagent. $1\text{ mole}$ of $\text{PbO}$ produces $1\text{ mole}$ of $\text{PbCl}_2$. Therefore, $0.029\text{ moles}$ of $\text{PbO}$ will produce $0.029\text{ moles}$ of $\text{PbCl}_2$.