In the following reaction,
$x M n_{2} O_{7} ⟶ y M n O_{2} + z O_{2}$ x, y, and z respectively are:

2,4,3

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

3,4,2

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

2,4,2

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

4,3,2

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

Open in App

Solution

The correct option is A
2,4,3

$M n_{2} O_{7}$ is splitted into two parts ( $+ 7 \times 2 M n_{2}$ and ${- 2 \times 7 O}_{7}$ )

Let the following equation be numbered ...(i)

Let the following equation be numbered ...(ii)

Now, multiply equation (i) by 28 and equation (ii) by 6 and add them as

$28 M n_{2} O_{7} + 12 M n_{2} O_{7} ⟶ 56 M n O_{2} + 42 O_{2}$ ...(iii)

In the above equation (iii), balance the $M n$ atoms disregarding $O$ atoms.

$40 M n_{2} O_{7} ⟶ 56 M n O_{2} + 24 M n O_{2} + 42 O_{2}$ ...(iv)

In the above equation, now balance the $O$ atoms to get

$40 M n_{2} O_{7} ⟶ 80 M n O_{2} + 42 O_{2} + 18 O_{2}$

The final, balanced equation is:

$2 M n_{2} O_{7} ⟶ 4 M n O_{2} + 3 O_{2}$

Suggest Corrections

Similar questions

In the following reaction,
$x M n_{2} O_{7} ⟶ y M n O_{2} + z O_{2}$ x, y, and z respectively are:

For the redox reaction,
$x C_{2} O_{4}^{- 2} + y M n O_{4}^{-} + z H^{+} \to y M n^{+ 2} + \frac{z}{2} H_{2} O + 2 x C O_{2}$
x, y and z respectively are:

Q.
If the balance reaction is as given below, find the value of

x, y

and

z

x M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) \to y M n O_{2} (s) + 2 H_{2} O + z O H^{-}

Q. Consider the following reaction,

x M n O_{4}^{-} + y C_{2} O_{4}^{2 -} + z H^{+}

\to

x M n^{2 +} + 2 y C O_{2} + \frac{z}{2} H_{2} O

The value's of

x, y

and

z

in the reaction are respectively:

Join BYJU'S Learning Program

In the following reaction, xMn2O7⟶yMnO2+zO2 x, y, and z respectively are:

In the following reaction,
$x M n_{2} O_{7} ⟶ y M n O_{2} + z O_{2}$ x, y, and z respectively are: