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Introduction to Qualitative Analysis

3.9g of a mix...

Question

3.9g of a mixture of Al and Al2O3 ,when reacted with a solution of sodium hydroxide produced 840 ml of a gas at NTP . find the % composition of the mixture

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Solution

Dear Student,

At first, let's figure out the reaction: Al₂O₃ does not react with NaOH. So, the reaction is:

$2 A l + 2 N a O H + 6 H_{2} O \to 2 N a A l (O H)_{4} + 3 H_{2} A t N T P a s 20 ° C a n d 1 a t m : n = \frac{P V}{R T} = \frac{(1 a t m) \times (0.840 L)}{((0.08205746 L a t m / K m o l) \times (20 + 273) K)} = 0.03494 m o l o f H_{2} T h i s c o r r e s p o n d s t o (0.03494 m o l H_{2}) \times (2 m o l A l / 3 m o l H_{2}) \times (26.98154 g A l / m o l) = 0.63 g A l A n d, (3.9 g t o t a l) - (0.63 g A l) = 3.3 g A l_{2} O_{3}$

Regards
Arnab Bhattacharya

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