Balance the following nuclear equations. #""_(0)^(1) n + ""_(92)^(235) U → ""_(36)^(92) Kr + ??? + 3 ""_(0)^(1) n + "energy"#

1 Answer
Truong-Son N.
Nov 20, 2017

#""_(0)^(1) n + ""_(92)^(235) U → ""_(36)^(92) Kr + ""_(56)^(141) Ba + 3 ""_(0)^(1) n + "energy"#

#""_(0)^(1) n + ""_(92)^(235) U → ""_(36)^(92) Kr + ??? + 3 ""_(0)^(1) n + "energy"#

Notice how the mass numbers and atomic numbers aren't balanced (and thus, the mass is not balanced). You can set this up with a system of equations (note that the neutrons contribute nothing to the atomic number and #1# to the mass number).

Let #A# be the mass number of the unknown isotope #X#, and #Z# be the atomic number of #X#.

This gives:

#1 + 235 = 92 + A + 3 cdot 1#

#0 + 92 = 36 + Z + 3 cdot 0#

The first equation gives the mass number to be:

#A = 236 - 92 - 3 = 141#

The second equation gives the atomic number to be:

#Z = 92 - 36 - 0 = 56#

The element written as #""_(56)^(141) "X"# is the barium-141 isotope, so the missing product is #color(blue)(""_(56)^(141) "Ba")#. As a result your balanced reaction is:

#""_(0)^(1) n + ""_(92)^(235) U → ""_(36)^(92) Kr + ""_(56)^(141) Ba + 3 ""_(0)^(1) n + "energy"#

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