4.0 g of a mixture containing Na_2CO_3 and N | vedclass

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(B) The chemical reaction for the heating of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$. $Na_2CO_3$ does not decompose at $673 \ K$.

Vedclass · Accepted Answer

$58$

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(B) The chemical reaction for the heating of $NaHCO_3$ is: $2NaHCO_3 ightarrow Na_2CO_3 + H_2O + CO_2$. $Na_2CO_3$ does not decompose at $673 \ K$. The loss in mass is due to the release of $H_2O$ and $CO_2$. The molar mass of $2NaHCO_3$ is $2 	imes 84 = 168 \ g/mol$. The mass of $H_2O + CO_2$ released is $18 + 44 = 62 \ g$. Let the mass of $NaHCO_3$ be $x \ g$. Loss in mass = $(62/168) 	imes x = 0.62 \ g$. $x = (0.62 	imes 168) / 62 = 1.68 \ g$. Mass of $Na_2CO_3 = 4.0 - 1.68 = 2.32 \ g$. Percentage of $Na_2CO_3 = (2.32 / 4.0) 	imes 100 = 58 \%$.

$4.0 \ g$ of a mixture containing $Na_2CO_3$ and $NaHCO_3$ is heated to $673 \ K$. The loss in mass of the mixture is found to be $0.62 \ g$. The percentage of sodium carbonate in the mixture is:

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