You visited us 1 times! Enjoying our articles? Unlock Full Access!

Limiting Reagent or Reactant

The reaction,...

Question

The reaction,

$O_{3} + N O ⟶ O_{2} + N O_{2}$

$0.740$ g of $O_{3}$ reacts with $0.670$ g of $N O$ to produce $N O_{2}$ . The limiting reagent is:

N O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

O_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Both are in equimolar ratio

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both are in stoichiometric ratio

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $O_{3}$
$O_{3} + N O ⟶ O_{2} + N O_{2}$

1 mole of $O_{3}$ reacts with 1 mole of $N O$ .

$0.74 g O_{3} = \frac{0.74}{48} = 0.0154 m o l O_{3}$

$0.67 g N O = \frac{0.67}{30} = 0.0223 m o l N O$

$O_{3}$ is the limiting reagent and $N O$ is in excess $= 0.0223 - 0.0154 = 0.007 m o l$

Hence, option B is correct.

Suggest Corrections

Similar questions

0.740

O_{3}

0.670

N O

N O_{2}

O_{3}, O_{2}, N O

N O_{2}

25^{o} C

N O (g) + N O_{2} (g) ⇌ N_{2} O_{3} (g); K_{P 1}

N O

N O_{2}

3 : 5

N O_{2}

0.5 a t m,

K_{P 1}

K O_{2}, N O_{2}, S O_{2}, B a O_{2}, C l O_{2}, O_{2}, N O, N_{2} O, O_{3}, C O

N O

O_{2}

N O_{2}

N O_{2}

O_{2}

Join BYJU'S Learning Program