Question

The volume of CO2 (9) evolved at STP on heating 20 g CaCO3 is (1 mol gas at STP = 22.4 L)
11.21 L
5.6L
4.48 L
2.24 L​

Answer 1

Answer:

ans is 4.48L

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Answer 2

Answer:

The correct answer is option (3) 4.48L.

When 20 g CaCO3 is heated at STP, 4.48L of CO2 is released.

Explanation:

Given :

20g of CaCO3

1 mol gas at STP = 22.4L

To find :

Volume of CO2 evolved at STP on heating 20g of CaCO3

Solution :

We know that, 100g = 1 mol

We know that, 100g = 1 mol Therefore, 20g = 1/5 mol

According to the question ,

1 mol gas at STP = 22.4L

Therefore, 1/5 mol= X

Considering the above two equations we come to a conclusion that -

X = 22.4 x 1/5

Hence, X = 4.48 L

According to our above calculations, we can say that the volume of CO2 evolved at STP on heating 20g of CaCO3 is 4.48L.

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