Question

What will be the volume of co2 liberated at ntp by heating 20 gram of 90% pure limestone csco 3?

Answer 1

Answer

Given ,

Mass of Mixture = 20g. [It contains impurities]

Purity of Limestone ( CaCO3 ) = 90%

Then , Amount of Limestone present in the mixture =

Then , Amount of Limestone present in the mixture =90/100 * 20 = 18g.

Now,

1 mole of CaCO3 = 40+12+3*16 [Ca=40;C=12;O=16]

= 100g

Now , No . of. moles in 18g CaCO3 = 18/100.

= 0.18 moles.

CaCO3 => CaO + CO2

If 1 mole CaCO3 gives 1 mole CO2

1 mole CaCO3 gives 1 mole CO2 Then , 0.18 moles of CaCO3 gives 0.18moles CO2

Since , at S.T.P Vol of Any Gas = 22.4 l or 22.4 dm^3

So , it contains 0.18 × 22.4 = 4.032 l or dm^3 of CO2.

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