Question

250.0 grams of uranium-235 are placed in a reactor. A nucleus of uranium-235 absorbs a neutron and undergoes nuclear fission to produce barium-141 and krypton-92. A single atom’s fission produces 211.3 MeV of energy.

Answer 1

Answer on Question #77995, Chemistry / General Chemistry

250.0 grams of uranium-235 are placed in a reactor. A nucleus of uranium-235 absorbs a neutron and undergoes nuclear fission to produce barium-141 and krypton-92. A single atom's fission produces 211.3 MeV of energy.

Solution

$${ } _ { 9 2 } ^ { 2 3 5 } U + { } _ { 0 } ^ { 1 } n \rightarrow { } _ { 5 6 } ^ { 1 4 1 } B a + { } _ { 3 6 } ^ { 9 2 } K r + { } _ { 0 } ^ { 1 } n$$

Find amount of substance of $250.0\mathrm{g}$ of uranium-235:

$$n = m / M;$$ $$n \left(^{235}_{92} U\right) = \frac{250.0\,g}{235\,g/mol} = 1.064\,mol$$

Find how many atoms of uranium-235 are in 1.064 mol:

$$N = N _ {A} \cdot n$$

Where $N_A$ is Avogadro number, $N_A = 6.02 \cdot 10^{23} \, \text{mol}^{-1}$

$n$ – amount of substance

$$N \left( \left(^{235}_{92} U \right) = 6.02 \cdot 10^{23} \cdot 1.064 = 6.404 \cdot 10^{23} \right)$$

Though there is no question in this task we can make an assumption that the total value of energy produced by $250.0\,\mathrm{g}$ of uranium -235 undergoing nuclear fission is asked.

A single atom's fission produces 211.3 MeV of energy. As we know the number of uranium-235 atoms $(6.404 \cdot 10^{23})$, we can find the energy produced by these atoms:

$$E = 211.3\,\mathrm{MeV} \cdot 6.404 \cdot 10^{23} = 1.35 \cdot 10^{26}\,\mathrm{MeV}$$

Answer: $1.35 \cdot 10^{26}\,\mathrm{MeV}$