Correct Answer - D
According to balanced equation,
`{:(PbO+,2HCl_((aq.))rarr,PbCl_(2)+,H_(2)O(l)),("1 mol","2 mol","1 mol","2 mol"):}`
Calculating moles of reactants :
`n_(PbO) = (6.5g)/(22.3g mol^(-1)) = 0.029 mol`
`n_(NCl) = (3.2g)/(36.5g mol^(-1)) = 0.87 mol`
Calculating moles of product formed from reactions:
`0.029 mol PbO xx (1 mol PbCl_(2)) = 0.029 mol PbCl_(2)`
`0.087 mol HCl xx (1 mol PbCl_(2))/(2 mol HCl) = 0.44 mol PbCl_(2)`
Thus, `PvbO` is the limiting reactant and `0.029 moln PbCl_(2)` will be formed.
