How many moles of lead (ll) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCI

$\mathrm{PbO}+2\mathrm{HCl}⟶{\mathrm{PbCl}}_2+{\mathrm{H}}_2\mathrm{O}$

Atomic mass of PbO = 207 + 16 = 223

Atomic mass of HCl = 36.5

Moles of PbO = $\frac{6.5}{223}$​ = 0.029 moles

Moles of HCl = $\frac{3.2}{36.5}$​ = 0.088 moles

From the above chemical equation, we see that PbO is the limiting reagent.

∴ 1 mole of PbO gives 1 mole of PbCl_2​

0.029 moles of PbO forms 0.029 moles of PbCl₂​.