Question

the volume of CO2 evolved at stp on heating 20 g of CaCo3​

Answer 1

Explanation:

100g = 1 mol

20g = 1/5 mol

1mol = 22.4 litter

1/5 mol = 22.4 * 1/5

= 4.48 litter

Answer 2

Answer:

The volume of $CO_{2}$ evolved at STP on heating 20g. of $CaCO_{3}$ is found to be 4.48 L​                  

Explanation:

Before solving the question, let us write down the data given:

• Weight of $CaCO_{3}$ = 20g.

• The volume of 1 mol. of gas at STP= 22.4 L

To be calculated, the volume of $CO_{2}$ evolved at STP on heating 20g. of $CaCO_{3}$ .

As we know that 1 mol.= 100g.

That means 100g.= 1 mol.

20g.=$\frac{1}{100}$ × $20$ = 1/5 mol.

• As per the question, 1 mol. of gas at STP= 22.4 L  

Hence to be calculated 1/5 mol. of gas= ?

• Let 1/5 mol. of gas= x

• So solving the above two equations we get,

x= $22.4$ × $\frac{1}{5}$

x= 4.48 L.

∴ x=4.48 L.

Hence the volume of $CO_{2}$ evolved at STP on heating 20g. of $CaCO_{3}$.​

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