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

Concentration Terms

What volume o...

Question

What volume of $O_{2}$ measured at standard conditions will be formed by the action of $100$ mL of $0.5 N$ $K M n O_{4}$ on hydrogen peroxide in an acidic solution?

0.12

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

0.28

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

0.56

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

1.12

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

Open in App

Solution

The correct option is B $0.56$ L
The reaction is as follows:
$2 K M n O_{4} + 5 H_{2} O_{2} + 3 H_{2} S O_{4} \to 2 M n S O_{4} + 5 O_{2} + K_{2} S O_{4} + 8 H_{2} O$
So, number of moles of $K M n O_{4} = 0.1$ (given)
So, $\frac{100 \times 0.1}{1000} = 0.01$ mol of $K M n O_{4}$ will produce $(\frac{5}{2}) \times 0.01 = 0.025$ mol of $O_{2} = 0.025 \times 22.4 = 0.56$ L.

Suggest Corrections

Similar questions

O_{2} (g)

K M n O_{4}

K M n O_{4} + H_{2} S O_{4} + H_{2} O_{2} \to K_{2} S O_{4} + M n S O_{4} + O_{2} + H_{2} O

O_{2} (g)

K M n O_{4}

K M n O_{4} + H_{2} S O_{4} + H_{2} O_{2} \to K_{2} S O_{4} + M n S O_{4} + O_{2} + H_{2} O

O_{2} (g)

K M n O_{4}

K M n O_{4} + H_{2} S O_{4} + H_{2} O_{2} \to K_{2} S O_{4} + M n S O_{4} + O_{2} + H_{2})

H_{2} O_{2}

H_{2} O_{2}

O_{2}

_{4}

_{4}

_{2}

_{4}

_{2}

_{2}

\to

_{2}

_{4}

_{4}

_{2}

_{2}

Join BYJU'S Learning Program