The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline, is

Question

The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline, is (A) 96500 C (B) 5790 C (C) 95700 C (D) 57900 C

Tardigrade Admin · Accepted Answer

C6H5NO2+6H++6e- xrightarrowC6H5NH2+2H2O 1 mol (= 123 g) nitrobenzene requires 6 moles of electrons = 6 x 96500 C charge ∴ 12.3g nitrobenzene will require charge =(6× 96500× 12.3/123) =6× 9650=57900 C

Q. The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline, is

96500 C

5790 C

95700 C

57900 C

$C_{6} H_{5} N O_{2} + 6 H^{+} + 6 e^{-} C_{6} H_{5} N H_{2} + 2 H_{2} O$ 1 mol (= 123 g) nitrobenzene requires 6 moles of electrons = 6 x 96500 C charge $∴$ 12.3g nitrobenzene will require charge $= \frac{6 \times 96500 \times 12.3}{123}$ $= 6 \times 9650 = 57900 C$

Q. The number of coulombs required to reduce 12.3 g of nitrobenzene to aniline, is

96500 C

5790 C

95700 C

57900 C

C6​H5​NO2​+6H++6e−​C6​H5​NH2​+2H2​O 1 mol (= 123 g) nitrobenzene requires 6 moles of electrons = 6 x 96500 C charge ∴ 12.3g nitrobenzene will require charge =1236×96500×12.3​ =6×9650=57900C

$C_{6} H_{5} N O_{2} + 6 H^{+} + 6 e^{-} C_{6} H_{5} N H_{2} + 2 H_{2} O$ 1 mol (= 123 g) nitrobenzene requires 6 moles of electrons = 6 x 96500 C charge $∴$ 12.3g nitrobenzene will require charge $= \frac{6 \times 96500 \times 12.3}{123}$ $= 6 \times 9650 = 57900 C$