Question

how much amount of CaCO3 in gram having purity 50 per cent produces 0.56 litre of CO2 at stp on heating?​

Answer 1

Volume of CO2 produced = 0.56L

Hence:-

Number of Moles of CO2 Produced will be as follows :-

$moles=\dfrac{volume\:given}{22.4}\\ \\ \\moles=\dfrac{0.56}{22.4}\\ \\ \\moles=0.025\:moles$

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$\boxed{\bold{\sf{\Large{CaCO_3 \xrightarrow{HEAT} CaO+CO_2}}}}$

$\rule{200}{2}$

According to Reaction:-

1 mole Of CO2 is produced when 1 Mole of CaCO3 is heated

Hence,

To produce 0.025 moles of CO2 we require 0.025 moles of CaCO3.

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Thus:-

Molar mass of CaCO3 = 100 grams

Mass of pure CaCO3 taken will be :-

$=100*0.025=2.5\:grams$

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Since the compound taken has 50% purity.

this means that only 2.5 gram of pure CaCO3 is present in it.

$\rule{200}{2}$

Let the amount o fcompound taken be x

so,

50% of x =2.5

This means :-

x = 2.5*2 = 5 grams.