19 g of a mixture containing NaHCO_3 and Na_ | vedclass

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(D) The decomposition reaction of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$. Let the mass of $NaHCO_3$ be $x \ g$ and $Na_2CO_3$ be

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$10.6$

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(D) The decomposition reaction of $NaHCO_3$ is: $2NaHCO_3 ightarrow Na_2CO_3 + H_2O + CO_2$. Let the mass of $NaHCO_3$ be $x \ g$ and $Na_2CO_3$ be $y \ g$. Given $x + y = 19$. Moles of $CO_2$ produced = $\frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$. From the reaction,$2 \ mol$ of $NaHCO_3$ produces $1 \ mol$ of $CO_2$. So,moles of $NaHCO_3 = 2 	imes 0.05 = 0.1 \ mol$. Mass of $NaHCO_3 = 0.1 \ mol 	imes 84 \ g/mol = 8.4 \ g$. Mass of $Na_2CO_3$ in the original mixture = $19 \ g - 8.4 \ g = 10.6 \ g$.

$19 \ g$ of a mixture containing $NaHCO_3$ and $Na_2CO_3$ on complete heating liberated $1.12 \ L$ of $CO_2$ at $STP$. The weight of the remaining solid was $15.9 \ g$. What is the weight (in $g$) of $Na_2CO_3$ in the mixture before heating?

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