Question

Complete and balance the following reactions (i) 92 $U^{235}{+}_o{n}^1\to \dots \dots .{+}_{52}$ Te $137{+}_{40}$ Zr97
(ii) 34 Se $86\to \dots \dots .+2_{-1}{e}^o$

Answer 1

Let Z be the atomic number and A mass number of the unknown atom X.

For (i): As atomic number and mass number is conserved in a reaction, so

$92+0=Z+52+40\Rightarrow Z=0$ and $235+1=A+137+97\Rightarrow A=2$

Thus, balance equation will be ${}_{92}{U}^{235}{+}_0{n}^1\to 2_0{n}^1{+}_{52}T{e}^{137}{+}_{40}Z{r}^{97}$

for (ii): $34=Z+2(-1)\Rightarrow Z=36$ and $86=A+0\Rightarrow A=86$

The unknown atom is an isotope of Krypton (Kr) and the balanced equation will be, ${}_{34}{U}^{86}{\to }_{36}K{r}^{86}+2_{-1}{e}^0$