Calculate the volume of CO_2 at NTP that can | vedclass

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(B) The balanced chemical equation for the thermal decomposition of $CaCO_3$ is:$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$From the stoichiometry,

Vedclass · Accepted Answer

$35.8$

Vedclass · Answer

(B) The balanced chemical equation for the thermal decomposition of $CaCO_3$ is:$CaCO_{3(s)} ightarrow CaO_{(s)} + CO_{2(g)}$From the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ produces $1 \ mol$ of $CO_2$ gas,which occupies $22.4 \ L$ at $NTP$.Given mass of $CaCO_3 = 200 \ g$.Since the purity is $80 \%$,the actual mass of pure $CaCO_3$ is:$200 \ g 	imes 0.80 = 160 \ g$.Number of moles of $CaCO_3 = \frac{160 \ g}{100 \ g/mol} = 1.6 \ mol$.Since $1 \ mol$ of $CaCO_3$ produces $22.4 \ L$ of $CO_2$,$1.6 \ mol$ of $CaCO_3$ will produce:$1.6 	imes 22.4 \ L = 35.84 \ L$.Rounding to one decimal place,the volume is $35.8 \ L$.

Calculate the volume of $CO_2$ at $NTP$ that can be obtained by heating $200 \ g$ of $80 \%$ pure $CaCO_{3(s)}$.

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