Question

Complete and balance for the following equations that are thought to take place in the stars.

(a)$\mathbf{2} {}_{\mathbf{6}}{}^{\mathbf{12}}\mathbf{C}\to \mathbf{?}\mathbf{+}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

(b) $\mathbf{?}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{1}}\mathbf{H}\to {}_{\mathbf{6}}{}^{\mathbf{12}}\mathbf{C}\mathbf{+}{}_{\mathbf{2}}{}^{\mathbf{4}}\mathbf{He}$

(c) $\mathbf{2}{}_{\mathbf{2}}{}^{\mathbf{3}}\mathbf{He}\to \mathbf{?}\mathbf{+}\mathbf{2}{}_{\mathbf{1}}{}^{\mathbf{1}}\mathbf{H}$

Short answer

The completeand balanced reactions that take place in the stars are given below.

(a) $\text{2} {}_{\text{6}}{}^{\text{12}}\text{C}\to {}_{\text{12}}{}^{\text{23}}\text{Mg +}{}_{\text{0}}{}^{\text{1}}\text{n}$

(b) ${}_{\text{7}}{}^{\text{15}}\text{N +}{}_{\text{1}}{}^{\text{1}}\text{H}\to {}_{\text{6}}{}^{\text{12}}\text{C +}{}_{\text{2}}{}^{\text{4}}\text{He}$

(c)$\text{2}{}_{\text{2}}{}^{\text{3}}\text{He}{\to }_{\text{2}}^{\text{4}}\text{He + 2}{}_{\text{1}}{}^{\text{1}}\text{H}$

Step-by-step solution

  Nuclear reactions

In nuclear reactions, a heavier nucleus decays to form lighter particles while lighter nuclei combine to form heavier nucleus. The nuclear reactions are balanced by equating the sum of mass numbers and atomic numbers of reactants with products. Normally nuclear reactions are accompanied with various kinds of emission. The most common type of emissions are alpha particles, beta particles, and gamma radiations.

Balancing reaction (a)

There are two ${}^{\text{12}}\text{C}$ present as reactants.

Total number of protons present = 12

Total number of neutrons present = 12

$\text{2} {}_{\text{6}}{}^{\text{12}}\text{C}\to \text{? +}{}_{\text{0}}{}^{\text{1}}\text{n}$

It is seen from the equations that a neutron is formed in the reaction. Therefore, the daughter nuclei formed should have atomic number 12 and mass number 23.

The balanced chemical equation is given below.

$\text{2} {}_{\text{6}}{}^{\text{12}}\text{C}\to {}_{\text{12}}{}^{\text{23}}\text{Mg +}{}_{\text{0}}{}^{\text{1}}\text{n}$

Balancing reaction (b)

We will follow the same principle to balance the second equation.

$\text{? +}{}_{\text{1}}{}^{\text{1}}\text{H}\to {}_{\text{6}}{}^{\text{12}}\text{C +}{}_{\text{2}}{}^{\text{4}}\text{He}$

The sum the mass number of products $=12+4=16$

The sum of atomic numbers of the products$=6+2=8$

Therefore, the mass number of the missing reactant $=16-1=15$

And, the atomic number of the missing reactant$=8-1=7$

The element is identified to be ${}_{\text{2}}{}^{\text{4}}\text{He}$.

Therefore, the balanced equation is given below.

$\text{2}{}_{\text{2}}{}^{\text{3}}\text{He}{\to }_{\text{2}}^{\text{4}}\text{He + 2}{}_{\text{1}}{}^{\text{1}}\text{H}$