A solid mixture weighing 5 g contains equal | vedclass

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(B) Let the number of moles of $Na_2CO_3$ be $x$ and $NaHCO_3$ be $x$.The molar mass of $Na_2CO_3 = (2 	imes 23) + 12 + (3 	imes 16) = 106 \ g/mol$.

Vedclass · Accepted Answer

$789.0$

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(B) Let the number of moles of $Na_2CO_3$ be $x$ and $NaHCO_3$ be $x$.The molar mass of $Na_2CO_3 = (2 	imes 23) + 12 + (3 	imes 16) = 106 \ g/mol$.The molar mass of $NaHCO_3 = 23 + 1 + 12 + (3 	imes 16) = 84 \ g/mol$.Total mass of the mixture: $106x + 84x = 5 \ g$.$190x = 5 \implies x = \frac{5}{190} = 0.026316 \ mol$.The reactions with $HCl$ are:$Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$ (Requires $2x$ moles of $HCl$)$NaHCO_3 + HCl ightarrow NaCl + H_2O + CO_2$ (Requires $x$ moles of $HCl$)Total moles of $HCl$ required $= 2x + x = 3x = 3 	imes 0.026316 = 0.078948 \ mol$.Volume of $0.1 \ M \ HCl$ required $= \frac{	ext{moles}}{	ext{molarity}} = \frac{0.078948}{0.1} = 0.78948 \ L$.Converting to $mL$: $0.78948 	imes 1000 = 789.48 \ mL \approx 789.0 \ mL$.

$A$ solid mixture weighing $5 \ g$ contains equal number of moles of $Na_2CO_3$ and $NaHCO_3$. This solid mixture was dissolved in $1 \ L$ of water. What is the volume (in $mL$) of $0.1 \ M \ HCl$ required to completely react with this $1 \ L$ mixture solution?

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