Question

When 0.01 kg of CaCO3 is decomposed the CO2 produced occupies a volume at S.T.P.:

Answer 1

Answer:

1 mole of CaCO3 produces = 1 mole of CO2. So 0.1 mole of CO2 produces= 0.1 mole of CO2. 0.1 mole of CO2 will occupy a volume of = 22.4 x 0.1= 2.24 Litres.

Answer 2

Answer:

2.24 L CO2 will be produced in the decomposition.

Explanation:

We know the reaction for decomposition of CO2 from CaCO3:

CaCO3(s) → CaO(s) + CO2(g) ... equation (1)

From equation (1);

We get 1 mole CaCO3(s) produces 1 mole CO2(g).

Molecular mass of CaCO3 = 100.0869 g/mol

So, 1 mole CaCO3 = 100 g CaCO3

Hence we get;

$100g \: CaCO3 = 1 \: mole \: CaCO3 \\ 0.1 \: kg \: CaCO3 = 1 \: mole \: CaCO3 \\ \\ 0.01 \: kg \: CaCO3 = 0.1 \: mole \: CaCO3$

So 0.1 mole CO2 will be produced in the decomposition of 0.1 mole CaCO3.

In S.T.P the volume of 1 mole CO2 = 22.4 L

So, we get volume of 0.1 mole CO2 in STP is = 2.24 L

Hence 2.24 L CO2 will be produced.