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

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

Solution:

The reaction between PbO and HCl is:

PbO + 2 HCl → PbCl₂+ H₂O

From the reaction it is clear that 1 mole PbO reacts with 2 mole of HCl.

Now, moles of PbO = ^{Given mass} / _{molar mass} = ^{6.5 g}/_{223 g mol^-1} = 0.029 mol

and moles of HCl = ^{Given mass} / _{molar mass} = ^{3.2 g}/_{36.5 g mol^-1} = 0.088 mol

0.029 mol PbO reacts with HCl = 2*0.029 = 0.0058 mol

HCl is present in excess, so PbO is limiting reactant.

Now, from reaction 1 mole PbO produce 1 mol PbCl₂ 

So, 0.029 mol PbO will produce 0.029 mol PbCl₂.

Hence PbCl₂ produced is 0.029 mol.

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