Consider the following nuclear reactions involving X and Y

$\begin{array}{l}X⟶Y{+}_2^4\mathrm{He}\\ Y{⟶}_8{\mathrm{O}}^{18}{+}_1{\mathrm{H}}^1\end{array}$

If both neutrons as well as protons in both the sides are conserved in nuclear reaction then moles of neutrons in 4.6 g of X:

When both the neutrons, and protons in both sides, are conserved in nuclear reaction thus, reactions obtained as,

${ }_{11}^{23}\mathrm{Na}{\to }_9^{19}\mathrm{F}{+}_2^4\mathrm{He}$

${ }_9^{19}\mathrm{F}{\to }_8^{18}\mathrm{O}{+}_1^1\mathrm{H}$

X is Na in which mass number A=23.

Now, it is known that,

Number of protons + Number of neutrons = Mass number A = 23

Number of neutrons = 23 - number of protons =23−11=12

1 mole Na=23 gm Na=12 moles of neutrons

4.6 gm Na= $=\frac{(4.6\times 12)}{23}$ =2.4 moles of neutrons