Question

4g of a mixture of CaCO3 and sand is treated with an excess of HCl and 0.88g of CO2 is produced. What is the percentage of CaCO3 in the mixture

Answer 1

Convert 0.88 g CO2 to CaCO3.
That's 0.88 x (molar mass CaCO3/molar mass CO2) = ?g CaCO3
Then % CaCO3 in sample = (?g CaCO3/mass sample)*100 = ?

Answer 2

44gms of CO2
is produced by (40+12+48)100gms of CaCO3
0.88gm of CO2
must be produced by (100/44)×0.88= 2gm of CaCO3
So the amount of pure CaCO3
present in 4gm of impure sample is 2gms.
So percentage of purity = (2/4) ×100= 50

Answer 3

it is right

Answer 4

It's right