The moles of lead (II) chloride that will be formed from a reaction between $6.5 \text{ g}$ of $\text{PbO}$ and $3.2 \text{ g}$ of $\text{HCl}$ are:

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

First, we calculate the number of moles of each reactant: Molar mass of $\text{PbO} = 207.2 + 16 = 223.2 \text{ g mol}^{-1}$. Number of moles of $\text{PbO} = \frac{6.5 \text{ g}}{223.2 \text{ g mol}^{-1}} = 0.0291 \text{ mol}$.

Molar mass of $\text{HCl} = 1 + 35.5 = 36.5 \text{ g mol}^{-1}$. Number of moles of $\text{HCl} = \frac{3.2 \text{ g}}{36.5 \text{ g mol}^{-1}} = 0.0876 \text{ mol}$.

From the stoichiometry of the balanced equation, $1 \text{ mole}$ of $\text{PbO}$ requires $2 \text{ moles}$ of $\text{HCl}$. Therefore, $0.0291 \text{ moles}$ 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, $\text{HCl}$ is in excess, making $\text{PbO}$ the limiting reagent .

The amount of product formed is entirely dependent on the limiting reagent. Since $1 \text{ mole}$ of $\text{PbO}$ yields $1 \text{ mole}$ of $\text{PbCl}_2$, $0.0291 \text{ moles}$ of $\text{PbO}$ will produce $0.0291 \text{ moles}$ (approximately $0.029 \text{ moles}$) of $\text{PbCl}_2$.