Asked by Axis

What volume of CO2 gas is produced when 20 gm of 20% pure CaCO3 is completely heated?
Answered by DrBob222
Do you know how to work a stoichiometry problem if you have the quantities. 20 g of 20% means you have 20*0.2 - 4 g CaCO3 to start.
Answered by Axis
Honestly, I didn't know. Thank you DrBob222
Answered by DrBob222
I hit the - sign instead of the = sign.

you have 20*0.2 = 4 g CaCO3 to start.
Answered by sandesh
896ml
Answered by SACHIN
can i get full solution?
Answered by Dipesh
896
Answered by Kanchan
Can I get full solution of this question. Please
Answered by samundra
You need to start with 20% of 20gm of CaCO3.

20% of 20gm of CaCO3 = 20 * 0.2 = 4 gm

Mass of CaCO3 at NTP = 100gm

Mass of CO2 at NTP = 44 gm.

CaCO3 → CaO + CO2 on heating.

So,

100gm of CaCO3 gives 44 gm of CO2.

4 gm of CaCO3 gives (44/100 * 4) = 1.76 gm CO2.

At NTP,

44 gm of CO2 → 22.4 litres

1.76 gm of CO2 → (22.4/44 *1.76) = 0.896 litres

So, 896 ml CO2 is produced.
Answered by Preeti rana
It is good
Answered by Sneha Niraula
896mL
Answered by Random Guy
Ans:- 896mL
soln,
CaCo3 → CaO + CO2 on heating
So, 1 mole of CaCo3 gives 1 mole of CO2
Now From Avagadro's Law, 1 mole of CO2 at NTP occupies 22.4L
100g of CaCo3 produces 22.4L of CO2
Since, 20% of 20 = 4 So,
4g of CaCo3 produces 4 x 22.4/100
=0.896L
=896ml.
Answered by Bot
Yes, your solution is correct.