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Standard XII

Chemistry

Stoichiometric Calculations

A mixture of ...

Question

A mixture of $N a H C O_{3}$ and $N a_{2} C O_{3}$ weighed .0235 g. The dissolved mixture was reacted with excess of $8 a (O H)_{2}$ to form $2.1028 g B a C O_{3}$ , by the following reactions:
$N a_{2} C O_{3} + B a (O H)_{2} \to B a C O_{3} + 2 N a O H$
$N a H C O_{3} + B a (O H)_{2} \to B a C O_{3} + N a O H + H_{2} O$
What was the percentage of $N a H C O_{3}$ in the original mixture ?

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Solution

Let x g of $N a H C O_{3}$ be present in the mixture.
Mass of $N a_{2} C O_{3}$ in the mixture $= (1.0235 - x) g$
Number of moles of $N a H C O_{3} = \frac{x}{84}$
Number of moles of $N a_{2} C O_{3} = \frac{(1.0235 - x)}{106}$
Number of moles of $B a C O_{3}$ = Number Of moles of $N a H C O_{3} +$ Number of moles of $N a_{2} C O_{3}$
$\frac{21028}{197} = \frac{x}{84} + \frac{(1.0235 - x)}{106}$
$x = 0.4122$
Amount of $N a H C O_{3} = 0.4122 g$
Percentage of $N a H C O_{3} = \frac{0.4122}{1.0235} \times 100 = 40.27$

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A mixture of NaHCO3 and Na2CO3 weighed .0235 g. The dissolved mixture was reacted with excess of 8a(OH)2 to form 2.1028gBaCO3, by the following reactions: Na2CO3+Ba(OH)2→BaCO3+2NaOHNaHCO3+Ba(OH)2→BaCO3+NaOH+H2O What was the percentage of NaHCO3 in the original mixture ?