15. How many moles of lead (Il) chloride will be formedand 3.2 g offro

1.	15. How many moles of lead (Il) chloride will be formedand 3.2 g offrom a reaction between 6.5g of PbOHCI?(1) 0.029(3) 0.333AIPMT (Prelims)-2008](2) 0.044(4) 0.011
Answer» 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

15. How many moles of lead (Il) chloride will be formedand 3.2 g offrom a reaction between 6.5g of PbOHCI?(1) 0.029(3) 0.333AIPMT (Prelims)-2008](2) 0.044(4) 0.011

Answer»

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