In a nuclear reaction, both the mass number (A) and the atomic number (Z) are conserved. The given reaction can be written as:

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{139}_{54}Xe + ^{94}_{38}Sr + x$

Applying the law of conservation of mass number (superscripts): $235 + 1 = 139 + 94 + A_x$ $236 = 233 + A_x \implies A_x = 3$

Applying the law of conservation of atomic number (subscripts): $92 + 0 = 54 + 38 + Z_x$ $92 = 92 + Z_x \implies Z_x = 0$

The product $x$ has a total mass number of 3 and an atomic number of 0. This corresponds to 3 neutrons ($3 \times ^{1}_{0}n$).