A detailed scientific illustration showing a chemical reaction process. On one side, a scale showing a measurement of 20 grams, with a pile of 20% pure CaCO3 (Calcium Carbonate). The Calcium Carbonate is then being heated in a volumetric flask over a Bunsen burner. On the other end, there is an diffusion of gas which symbolises the CO2 gas being produced from the reaction. All components are placed on a laboratory table with a sterile, clinical feel to room and the lighting.

What volume of CO2 gas is produced when 20 gm of 20% pure CaCO3 is completely heated?

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.

896ml

Can I get full solution of this question. Please

can i get full solution?

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.

What volume of CO2 gas is produced when 20 gm of 20% pure CaCO3 is completely heated?

You need to start with 20% of 20gm of CaCO3.

896ml

Can I get full solution of this question. Please

can i get full solution?

Ans:- 896mL

It is good

896

Yes, your solution is correct.

I hit the - sign instead of the = sign.

896mL