Question

How many moles of lead (II) chloride will be formed
from a reaction between 6.5g of PbO and 3.2 g of
HCI?
[AIPMT (Prelims)-2008)​

Answer 1

Explanation:

The equation for the reaction between PbO and HCl is as follows:

PbO + 2HCl------------> PbCl2  + Cl + H2O

Therefore 1 moles of PbO require 2 moles of HCl to give 1 mole of PbCl2.

Calculate the reagent, between HCl and PbO, that is the limiting reagent.

Pb-207.2  Cl-35.5  O-16  H-1

moles=mass/molar mass

moles of PbO  = 6.5/(207+16)  = 6.5/223.2  = 0.029148

if 1 mole = 2 moles of HCl 

then 0.029148= 2x0.029148= 0.058296

moles of HCl= 3.2/(1+35.5) = 3.2/36.5  =  0.08767

if 2 moles of HCl=  1 mole of PbO

then 0.08767= 1x 0.8767/2  =  0.04384

Therefore PbO is the limiting reagent.

Calculate moles of PbCl2 produced as follows:

if 1 mole of PbO  = 1 mole of PbCl2

then 0.029148 will give=  1 x 0.029148 moles

Therefore moles of PbCl2 produced = 0.029148 moles

Hope it helps you....

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