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Stoichiometric Calculations

124 g of mixt...

Question

$124$ g of mixture containing $N a H C O_{3}$ , $A l C l_{3}$ and $K N O_{3}$ requires $500$ mL, $8$ % $(w / w)$ $N a O H$ solution $[d_{N a O H} = 1.8 g / m L]$ for complete neutralization. On heating same amount of mixture, it shows loss in weight of $18.6$ g. Weak base formed does not interfere in reaction. Assume $K N O_{3}$ does not decompose under given conditions. The $%$ (nearest integer) of $A l C l_{3}$ in the mixture is ________ .

$A l C l_{3} + 3 N a O H \to A l {(O H)}_{3} + 3 N a C l$
$N a H C O_{3} + N a O H \to N a_{2} C O_{3} + C O_{2} + H_{2} O$

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Solution

The loss in mass on heating is 18.6 g.

$2 N a H C O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$ .

18.6 g corresponds to $\frac{18.6}{44 + 18} = 0.3$ moles of $C O_{2} + H_{2} O$ which are obtained from 0.6 moles (50.4 g) of sodium hydrogen carbonate.

500 ml, 8% w/w NaOH of density 1.8 g/ml corresponds to $500 m l \times 1.8 g / m l = 900 g \times \frac{8}{100} = 72 g N a O H$ .

This corresponds to $\frac{72}{40} = 1.8 m o l e s$ .

1-mole sodium hydrogen carbonate will react with 1 mole of NaOH.

Hence, 0.6 moles of $N a H C O_{3}$ will react with 0.6 mole of NaOH.

1 mole $A l C l_{3}$ will react with 3 moles of NaOH.

The number of moles of $A l C l_{3} = \frac{1.8 - 0.6}{3} = 0.4 m o l$ .

Mass of $A l C l_{3} = 0.4 \times 133.3 = 53.3 g$

Percentage of $A l C l_{3} = \frac{53.36}{124} \times 100 = 43$ %

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