Question

By passing electric current, $NaCl{O}_3$ is converted into $NaCl{O}_4$ according to the following equation:
$NaCl{O}_3+H_{2}O\to NaCl{O}_4+H_2$
How many moles of $NaCl{O}_4$ will be formed when three Faradays of charge is passed through $NaCl{O}_3$?

• A. 0.75
• B. 1
• C. 1.5
• D. 3

Answer 1

The correct option is C 1.5

$NaCl{O}_3+H_{2}O\to NaCl{O}_4+H_2$
(+5) (+7)
2 F charge is required for 1 mole conversion.
$\therefore$ 3 F of charge = $\frac{3}{2}$=1.5 moles of the compound are converted and .5 moles of $NaCl{O}_4$ will be formed.