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Mole Concept and Avogadro's Number

47 g of impur...

Question

47 g of impure al2o3 is reduced electrotytically to give aluminimum and oxygen , calculate the amount of aluminimum produced

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Solution

$2 A l_{2} O_{3} \to 4 A l + 3 O_{2} 2 (2 \times 27 + 3 \times 16) - - - - - > 4 \times 27 204 g o f A l u m i n i u m o x i d e p r o d u c e s 108 g o f A l 47 g o f A l u m i n i u m o x i d e p r o d u c e s = \frac{108}{204} \times 47 = 24.88 g o f A l$

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Q. The electrolyte used in electrolytic refining of impure aluminium is __________.

Q. Assertion :Al obtained only by electrolytic reduction of Alumina (pure) Reason: Aluminium metal has a strong affinity for oxygen and the oxide of Al is

A l_{2} O_{3}

, which is a very stable oxide. Hence ordinary chemically reducing agents are not sufficient to reduce

A l_{2} O_{3}

Q. The formula of Aluminium oxide is

A l_{2} O_{3}

. Find the valencies of Aluminium and Oxygen.

A l C l_{3}

reacts with 120 g of NaOH to give aluminium hydroxide. The produced aluminium hydroxide reacts with 3 mol of

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to give aluminium sulphate. Calculate the moles of aluminium sulphate produced in the reaction.

A l C l_{3} + 3 N a O H \to A l (O H)_{3} + 3 N a C l

2 A l (O H)_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 6 H_{2} O

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reacts with 120 g of NaOH to give aluminium hydroxide. The produced aluminium hydroxide reacts with 3 mol of

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A l C l_{3} + 3 N a O H \to A l (O H)_{3} + 3 N a C l

2 A l (O H)_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 6 H_{2} O

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