What volume of CO_2 at STP is obtained by the | vedclass

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(D) The balanced chemical equation for the thermal decomposition of potassium bicarbonate is:$2KHCO_{3(s)} 	o K_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$M

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(D) The balanced chemical equation for the thermal decomposition of potassium bicarbonate is:$2KHCO_{3(s)} 	o K_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$Molar mass of $KHCO_3 = 39 + 1 + 12 + (3 	imes 16) = 100 \ g/mol$.Number of moles of $KHCO_3 = \frac{20 \ g}{100 \ g/mol} = 0.2 \ mol$.From the stoichiometry of the reaction,$2 \ mol$ of $KHCO_3$ produces $1 \ mol$ of $CO_2$.Therefore,$0.2 \ mol$ of $KHCO_3$ produces $\frac{0.2}{2} = 0.1 \ mol$ of $CO_2$.At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.Volume of $CO_2 = 0.1 \ mol 	imes 22.4 \ L/mol = 2.24 \ L$.

What volume of $CO_2$ at $STP$ is obtained by thermal decomposition of $20 \ g$ $KHCO_3$? [Atomic weight of $K = 39$]
$2KHCO_{3(s)} \to K_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$

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What volume of $CO_2$ at $STP$ is obtained by thermal decomposition of $20 \ g$ $KHCO_3$? [Atomic weight of $K = 39$]$2KHCO_{3(s)} \to K_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$