Correct Answer - B
`underset("x mole")(PbO)+underset("2x moles")(2HCl)rarrunderset("x mole")(PbCl_(2))+H_(2)O`
`underset("= 0.029 mole")((6.5)/(224)"mole")" "underset("= 0.087 mole")((3.2)/(36.5)"mole")`
Thus, 0.029 mole of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl.