English, asked by ashwinighumbre99, 9 months ago

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

Answers

Answered by langlenchanu22
2

Explanation:

100g = 1 mol

20g = 1/5 mol

1mol = 22.4 litter

1/5 mol = 22.4 * 1/5

= 4.48 litter

Answered by pragyan07sl
1

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}.​

#SPJ3  

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