The volume of oxygen liberated from 0.68 gm of $H_{2} O_{2}$ is

112 ml

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

224 ml

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

56 ml

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

336 ml

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

Open in App

Solution

The correct option is B 224 ml
We know that

$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$
$2 \times 34 g$ $22400 m l$

$∵ 2 \times 34 g = 68 g$ of $H_{2} O_{2}$ liberates 22400 ml $O_{2}$ at STP

$∴$ 0.68 g of $H_{2} O_{2}$ liberates $= \frac{0.68 \times 22400}{68} = 224 m l$

Suggest Corrections

Similar questions

15

20

H_{2} O_{2}

15 mL

20

H_{2} O_{2}

100

15 V

H_{2} O_{2}

Join BYJU'S Learning Program