The correct option is A 2.016 litres #10; #10; #10; #10; #9; #10; #9; #10; #9; #10; #9; #10; #10; #10;Note that t ...

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Concentration Terms

CaCO3 is 90

Question

$C a C O_{3}$ is $90$ % pure. Volume of $C O_{2}$ collected at STP when $10 g$ of $C a C O_{3}$ is decomposed is:

2.016 l i t r e s

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1.008 l i t r e s

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10.08 l i t r e s

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20.16 l i t r e s

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Solution

The correct option is A $2.016 l i t r e s$

Note that the gas produced is $C O_{2}$
Equation:
$C a C O_{3} \to C a O + C O_{2}$
1 mol CaCO $_{3}$ produces 1 mol $C O_{2}$

You have $\frac{90}{100} \times 10 = 9 g$ pure $C a C O_{3}$

Molar mass $C a C O_{3}$ = $100 g / m o l$
$9 g C a C O_{3} = 9100 = 0.09 m o l C a C O_{3}$
This will produce 0.09 mol $C O_{2}$
At STP $1 m o l = 22.4 L$
$0.09 m o l = 0.09 \times 22.4 = 2.016 L$

Hence the volume of the $C O_{2}$ collected at STP when is decomposed is = $2.0 L$

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