Question

In the equation
xp + HNO3 → HPO3 + yNO + H2O
(a) x = 5, y = 5
(b) x = 5, y = 3
(c) x = 3, y = 5
(d) x = 3, y = 3
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Answer 1

AnSwer :-

Option (c) x = 3, y = 5 is correct.

Explaination :-

By using Oxidation Number method

Let's find x and y in steps.

Step - 1.

P + HNO$_{3}$ $\longrightarrow$ HPO$_{3}$ + NO + H$_{2}$O

Step - 2.

Writing the oxidation numbers of the elements.

$\sf P + HNO_{3}\longrightarrow HPO_{3}+ NO +H _{2}O$

0 ⠀⠀+5 ⠀⠀⠀⠀⠀+5 ⠀⠀ ⠀+2

Thus here P is oxidized (0 to +5) and N in HNO$_{3}$ is reduced (+5 to +2) .

Now, writing separate equation for the reducing agent (P) and oxidising agent (N in HNO$_{3}$).

Oxidation :- P⁰ $\longrightarrow$ P$^{+5}$ ↑ 5

Reduction :- N$^{+5}$ $\longrightarrow$ N $^{+2}$ ↓3

Step - 3.

Adding electrons according to change in oxidation number.

P⁰ $\longrightarrow$ P$^{+5}$ + 5e$^{-}$ ⠀⠀⠀⠀⠀⠀⠀⠀......(i)

N$^{+5}$ + 15e$^{-}$ $\longrightarrow$ 5N$^{+2}$ ⠀⠀⠀⠀⠀⠀⠀.....(ii)

Step - 4.

Multiply equation (i) by 3 and (ii) by 5.

3P⁰ $\longrightarrow$ 3P$^{+5}$ + 15e$^{-}$ ⠀⠀⠀⠀⠀⠀⠀.....(iii)

5N$^{+5}$ + 15e$^{-}$ $\longrightarrow$ 5N$^{+2}$ ⠀⠀⠀⠀⠀⠀.....(iv)

Adding (iii) and (iv)

3P⁰ + 5N$^{+5}$ $\longrightarrow$ 3P$^{+5}$ + 5N$^{+2}$

Therefore, the required reaction can be written as

+ 5HNO$_{3}$ $\longrightarrow$ 3HPO$_{3}$ + 5NO

Hence, the required Balanced equation can be represented as

3P + 5HNO$_{3}$ $\longrightarrow$3HPO$_{3}$+ 5NO + H$_{2}$O.

Thus, x = 3 and y = 5