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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0.5

0.25

answer is B.

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