Question

$A{l}_2{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0\times {10}^4$ 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 $Al=27gmo{l}^{-1})$

• A. $1.3\times {10}^{4}g$
• B. $9.0\times {10}^{3}g$
• C. $8.1\times {10}^{4}g$
• D. $2.4\times {10}^{5}g$

Answer 1

The correct option is C $8.1\times {10}^{4}g$

$W=z\times I\times tz=\frac{molarmass}{nf\times 96500}z=\frac{27}{3\times 96500}W=\frac{27}{3\times 96500}\times 4\times {10}^4\times 6\times 60\times 60=8.1\times {10}^{4}g$