Question

If 20gm of CaCO3is treated witj 20gm of HCL How MANY GM OF CO2 BE PRODUCED

Answer 1

Given,
CaCO3(s)+ 2HCl(aq)→ CaCl2(aq)+ CO2(g)+ H2O(l)

In this equation 1 mole of CaCO3 reacts with
2 moles of HCL

Number of moles of CaCO3=
(20 g) / (100 g/mol)= 0.2 moles

Ratio of CaCO. 3:HCl=1:2

Moles of HCl required = 0.2 x 2 = 0.4

acctual Number of moles of HCl =
(20 g) / (36.6 g/mol)
=0.55 moles

Therefore HCl is in execss
and hence
CaCO3is the limiting reagentMoles of
CO2= 0.2 moles
Hence
,Mass of CO2= 0.2 mol x 44 g/mol
=8.8 g

Answer 2

CaCO3+2HCl-------------------=>CaCl2+H2O+CO2
1 mole+2mole-----------------------------------=>1mole
100 g. +77 g ----------------------------------=>44 g
20 g. +20 g------------------------------------=>?

No. of moles of CaCO3 =20/100=0.2 moles
Ratio of CaCO3 :HCl=1:2
So, no. of moles of HCl=0.2×2=0.4 moles
Actual no. of moles=20/36.5=0.55 moles
HCl is in excess so we need to compare CaCO3 and CO2.

100 g of CaCO3 gives 44 g of CO2
20 g of CaCO3 gives how many grams of CO2
=20×44/100
=8.8 g

hope it helps you