Al 2 O 3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al 2 O 3 for 6 hours, what mass of aluminium is produced ? (Assume 100 % current efficiency, at. mass of Al =27 g mol -1 )

Question

Al 2 O 3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al 2 O 3 for 6 hours, what mass of aluminium is produced ? (Assume 100 % current efficiency, at. mass of Al =27 g mol -1 ) (A) 9.0 × 103 g  (B) 8.1 × 104 g  (C) 2.4 × 105 g  (D) 1.3 × 104 g

Tardigrade Admin · Accepted Answer

W = z × I × t z =( text molar mass / nf × 96500) z =(27/3 × 96500) W =(27/3 × 96500) × 4 × 104 × 6 × 60 × 60 =8.1 × 104 g

Q. $A l_{2} O_{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 1 0^{4}$ amperes of current is passed through molten $A l_{2} O_{3}$ for $6$ hours, what mass of aluminium is produced ?
(Assume $100%$ current efficiency, at. mass of $A l = 27 g m o l^{- 1}$ )

$9.0 \times 1 0^{3} g$

$8.1 \times 1 0^{4} g$

$2.4 \times 1 0^{5} g$

$1.3 \times 1 0^{4} g$

$W = z \times I \times t$
$z = \frac{molar mass}{n f \times 96500}$
$z = \frac{27}{3 \times 96500}$
$W = \frac{27}{3 \times 96500} \times 4 \times 1 0^{4} \times 6 \times 60 \times 60$
$= 8.1 \times 1 0^{4} g$

Q. Al2​O3​ is reduced by electrolysis at low potentials and high currents. If 4.0×104 amperes of current is passed through molten Al2​O3​ for 6 hours, what mass of aluminium is produced ? (Assume 100% current efficiency, at. mass of Al=27gmol−1 )

9.0×103g

8.1×104g

2.4×105g

1.3×104g