$224 c m^{3}$ of ammonia undergoes catalytic oxidation in presence of $P t$ to give nitric oxide and water vapour. The volume of oxygen required for the reaction is:
(All volumes measured at room temperature and pressure)

896 c c . o f O_{2}

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224 c c . o f O_{2}

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180 c c . o f O_{2}

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280 c c . o f O_{2}

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Solution

The correct option is D $280 c c . o f O_{2}$
Reaction:
$4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$
$4$ volumes of ammonia reacts with $5$ volumes of oxygen to give $4$ volumes of $N O$ .
So to react with $224$ $c m^{3}$ ammonia, oxygen requires is $5 \times 224 / 4$ = $280$ $c m^{3}$

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224cm3 of ammonia undergoes catalytic oxidation in presence of Pt to give nitric oxide and water vapour. The volume of oxygen required for the reaction is:(All volumes measured at room temperature and pressure)

The correct option is D 280 cc.of O2Reaction:4NH3+5O2⟶4NO+6H2O4 volumes of ammonia reacts with 5 volumes of oxygen to give 4 volumes of NO.So to react with 224 cm3 ammonia, oxygen requires is 5×224/4 = 280 cm3

$224 c m^{3}$ of ammonia undergoes catalytic oxidation in presence of $P t$ to give nitric oxide and water vapour. The volume of oxygen required for the reaction is:
(All volumes measured at room temperature and pressure)

The correct option is D $280 c c . o f O_{2}$
Reaction:
$4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$
$4$ volumes of ammonia reacts with $5$ volumes of oxygen to give $4$ volumes of $N O$ .
So to react with $224$ $c m^{3}$ ammonia, oxygen requires is $5 \times 224 / 4$ = $280$ $c m^{3}$