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Q.

Copper metal reacts with concentrated HNO₃ as Cu + 4HNO₃ →Cu(NO₃)₂ + 2NO₂ + 2H₂O . The number of moles of the HNO₃ reduced by one mole of the Cu will be

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a

2

b

0.25

c

4

d

0.5

answer is B.

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Detailed Solution

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$\overset{0}{Cu} + H \overset{+ 5}{N} O_{3} \to \overset{+ 2}{Cu} {({NO}_{3})}_{2} + \overset{+ 4}{N} O_{2} + H_{2} O$

Change in ox. state of 'Cu'=2

Change in ox. state of 'N' =1

Hence, 2moles of HNO₃ reduced by one mole of the Cu

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