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Stoichiometry

3.90 g of a m...

Question

$3.90$ g of a mixture of $A l$ and ${A l}_{2} O_{3}$ , when reacted with a solution of $N a O H$ produced $840$ mL of gas at NTP. Find $%$ of $A l_{2} O_{3}$ in mixture (as nearest integer).

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Solution

The reactions that are involved are:
$2 A l (s) + 2 N a O H + 6 H_{2} O \to 2 N a^{+} + 2 [A l (O H)_{4}]^{-} + 3 H_{2} (g)$
$2$ moles $3$ moles gas
$A l_{2} O_{3} + 2 N a O H + 3 H_{2} O \to 2 N a^{+} + 2 [A l (O H)_{4}]^{-}$
So, mass of $A l$ in sample $= \frac{2 \times 27 \times 840}{3 \times 22400} = 0.675$ g
So, $%$ of $A l$ $= \frac{0.675 \times 100}{3.9} = 17.3$ and
$%$ of $A l_{2} O_{3}$ in mixture $= 83$

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