Copper metal rects with conctrated HNO₃ as $\large Cu{\text{ }} + {\text{ }}4HN{O_3}{\text{ }}\xrightarrow{{\;\;\Delta \;\;}}Cu{\left( {N{O_3}} \right)_2}{\text{ }} + {\text{ }}2N{O_2}{\text{ }} + {\text{ }}2{H_2}O{\text{ }}$ . The number of moles of the HNO₃ reduced by one mole of the Cu will be

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answer is B.

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Detailed Solution

$\large Cu{\text{ }} + {\text{ }}4HN{O_3}{\text{ }}\xrightarrow{{\;\;\Delta \;\;}}Cu{\left( {N{O_3}} \right)_2}{\text{ }} + {\text{ }}2N{O_2}{\text{ }} + {\text{ }}2{H_2}O{\text{ }}$
$\large \begin{array}{*{20}{c}} {\mathop {Cu}\limits_O \; \to \;\mathop {C{u^{ + 2}}}\limits_{ + 3} } \\ { increase \;in\;oxidation\;number\; = \;2} \end{array}\left| {\begin{array}{*{20}{c}} {\mathop {NO_3^ - }\limits_{ + 5} \to \mathop {N{O_2}}\limits_{ + 4} } \\ { decrease \;in\;oxidation\;number\; = \;1} \end{array}} \right.$
Criss-cross method
1 mole of Cu reduces 2mole of $\large NO_3^ -$ ions