Question

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?
(A) 0.011
(B) 0.029
(C) 0.333
(D) 0.044

Answer 1

Answer:

(B) $0.029 mol PbCl_{2}$

Explanation:

As per the question,

Firstly, write the balanced reaction of lead oxide (PbO) with hydrogen chloride ( HCl), we get:

$PbO + 2HCl = PbCl_{2} + H_{2}O$

According to question,

6.5g PbO x 1mol PbO/223 g = 0.03 mol PbO

3.2g HCl x 1mol HCl/36.5g = 0.088 mol HCl

From this balanced equation we can say that PbO is the limiting reactant. Therefore, we will use PbO to find out the amount of moles of $PbCl_{2}$.

That is,

0.03 mol PbO x 1mol PbCl2/1molPbO = 0.029 mol PbO.

Hence, moles of lead (II) chloride formed = $0.029 mol PbCl_{2}$