QJEEM23Ch8aca: JEE Main 2023 Chemistry - Class 11 JEEM23Ch8aca Solution: See the following chemical reaction: The sum of and is…

To find the sum of X, Y, and Z, we need to balance the given chemical reaction:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow \mathrm{YCr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}$

Let's balance the reaction step by step:

1. Balance the chromium atoms:

There are 2 chromium atoms on the left side of the equation, so we need 2 chromium atoms on the right side. Thus, Y = 2.

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}$

2. Balance the oxygen atoms:

There are 7 oxygen atoms on the left side of the equation, so we need 7 oxygen atoms on the right side. Since each water molecule (H₂O) contains one oxygen atom, Z = 7.

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$

3. Balance the hydrogen atoms and charges:

There are 14 hydrogen atoms on the right side of the equation, so we need 14 hydrogen atoms on the left side. Thus, X = 14.

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$

Now the chemical reaction is balanced:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$

The sum of X, Y, and Z is:

X + Y + Z = 14 + 2 + 7 = 23

So, the sum of X, Y, and Z is 23.