4-5 g of aluminium -at- mass 27 amu- is deposited -160-at cathode from - Al-3- solution by a certain -160-quantity of electric charge- The volume of -160-hydrogen produced at STP from - H- ions in -160-solution by the same quantity of electric charge -160-will be-44-8L11-2L22-4L5-6L

The correct option is D 5.6LNo. of moles of Aluminium deposited =4.527=16 moles 2Al⟶ 2Al^3++6e^ 6H^++6e^ ⟶ 3H_2 2Al+6H^+⟶ 3H_2+2Al^3 ...

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Standard XII

Chemistry

Mole Concept

4.5 g of alum...

Question

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from $A l^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be:

44.8L

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11.2L

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22.4L

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5.6L

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Solution

The correct option is D 5.6L
No. of moles of Aluminium deposited $= \frac{4.5}{27} = \frac{1}{6}$ $m o l e s$

$2 A l ⟶ 2 A l^{3 +} + 6 e^{-}$

$6 H^{+} + 6 e^{-} ⟶ 3 H_{2}$

$⟹ 2 A l + 6 H^{+} ⟶ 3 H_{2} + 2 A l^{3 +}$

$2$ $m o l e s$ of Aluminium produce $3$ $m o l e s$ of hydrogen.
So, $2$ $m o l e s$ of $A l ⟶ 3$ $m o l e s$ of $H_{2}$

$\frac{1}{6}$ $m o l e s$ of $A l ⟶ \frac{1}{4}$ $m o l e s$ of $H_{2}$ .

So, $1$ $m o l e$ of $H_{2} ⟶ 22.4$ $L i t r e$

$\frac{1}{4}$ $m o l e$ of $H_{2} = \frac{22.4}{4} = 5.6$ $L$

Hence, the correct option is $D$

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4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge will be:

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from $A l^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be: