When aluminum oxide ( Al 2 O 3) is electrolyzed for the production of aluminum metal. For a given quantity of electricity, the number of moles of aluminum obtained if the volume of O 2 gas obtained is 201.6 liter measured at NTP is:

Question

When aluminum oxide ( Al 2 O 3) is electrolyzed for the production of aluminum metal. For a given quantity of electricity, the number of moles of aluminum obtained if the volume of O 2 gas obtained is 201.6 liter measured at NTP is: (A) 9 (B) 12 (C) 6 (D) 4.5

Tardigrade Admin · Accepted Answer

Number of equivalent of oxygen =(201.6/5.6)=36 (Equivalent volume of oxygen =5.6 liter at N.T.P.) ∴ Number of equivalent of Al =36 Mass of aluminum =36 × 9 g Number of moles of aluminum =(36 × 9/27)=12

Q. When aluminum oxide $(A l_{2} O_{3})$ is electrolyzed for the production of aluminum metal. For a given quantity of electricity, the number of moles of aluminum obtained if the volume of $O_{2}$ gas obtained is $201.6$ liter measured at NTP is:

9

12

6

4.5

Number of equivalent of oxygen $= \frac{201.6}{5.6} = 36$
(Equivalent volume of oxygen $= 5.6$ liter at N.T.P.)
$∴$ Number of equivalent of $A l = 36$
Mass of aluminum $= 36 \times 9 g$
Number of moles of aluminum $= \frac{36 \times 9}{27} = 12$

Q. When aluminum oxide (Al2​O3​) is electrolyzed for the production of aluminum metal. For a given quantity of electricity, the number of moles of aluminum obtained if the volume of O2​ gas obtained is 201.6 liter measured at NTP is:

9

12

6

4.5

Number of equivalent of oxygen =5.6201.6​=36 (Equivalent volume of oxygen =5.6 liter at N.T.P.) ∴ Number of equivalent of Al=36 Mass of aluminum =36×9g Number of moles of aluminum =2736×9​=12

Q. When aluminum oxide $(A l_{2} O_{3})$ is electrolyzed for the production of aluminum metal. For a given quantity of electricity, the number of moles of aluminum obtained if the volume of $O_{2}$ gas obtained is $201.6$ liter measured at NTP is:

Number of equivalent of oxygen $= \frac{201.6}{5.6} = 36$
(Equivalent volume of oxygen $= 5.6$ liter at N.T.P.)
$∴$ Number of equivalent of $A l = 36$
Mass of aluminum $= 36 \times 9 g$
Number of moles of aluminum $= \frac{36 \times 9}{27} = 12$