When 10 g of 90 % pure limestone is heated, | vedclass

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(B) The decomposition reaction of $CaCO_3$ is:$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$At $	ext{STP}$, heating $1 \ mol$ of $CaCO_3$ (

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$2.0$

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(B) The decomposition reaction of $CaCO_3$ is:$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$At $	ext{STP}$, heating $1 \ mol$ of $CaCO_3$ (i.e.,$100 \ g$) liberates $1 \ mol$ or $22.4 \ L$ of $CO_2$.Mass of pure $CaCO_3$ in $10 \ g$ of $90 \%$ pure limestone $= \frac{10 	imes 90}{100} = 9 \ g$.Since $100 \ g$ of $CaCO_3$ produces $22.4 \ L$ of $CO_2$,$9 \ g$ of $CaCO_3$ will liberate $= \frac{9 	imes 22.4}{100} \ L = 2.016 \ L \approx 2.0 \ L$ of $CO_2$.

When $10 \ g$ of $90 \%$ pure limestone is heated, the approximate volume $($in $L$ $)$ of $CO_2$ liberated at $\text{STP}$ is:

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