Question

The weight of lime obtained by heating 200 kg of 95% pure lime stone is :
(1) 98.4 kg
(2) 106.4 kg
(3) 112.8 kg
(4) 122.6 kg
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Answer 1

Answer :

The weight of lime obtained by heating 200 kg of 95% pure lime stone is :

(1) 98.4 kg

(2) 106.4 kg ✔️

(3) 112.8 kg

(4) 122.6 kg

SOLUTION :-

$\rm{CaCo_3 \longrightarrow CaO + CO_2}$

According to the question, it is given that there is 95% pure limestone ,then

Mass of $CaCo_3$ after heating = 200×

$Mass \: of \: CaCo_3 \: after \: heating = 200 \times \frac{25}{100} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 190 \: kg$

Now,

Molar mass of CaCO3 = [40 + 12+16×3]

Molar mass of CaCO3 = 100 g

And,

Molar mass of Lime (CaO) = [40+16]

Molar mass of lime (CaO) = 56 g

Hence,

For 100 g of CaCO3 we get 56 g of lime.

For 1 g of CaCO3 we get 0.56 g of lime.

Then, for 190 kg of CaCO3 we get 190×0.56 = 106.4 kg of lime.