Question

(_mathrm{zn}+ _mathrm{hno}_{3}
ightarrow _mathrm{zn}(mathrm{no}_{3})_{2}+ _mathrm{h}_{2})

Explanation:

Step1: Balance zinc atoms

On the left - hand side, we have 1 zinc atom in \(Zn\) and on the right - hand side, we have 1 zinc atom in \(Zn(NO_3)_2\). To balance zinc, if we put a coefficient of 3 in front of \(Zn\) and 3 in front of \(Zn(NO_3)_2\), we have 3 zinc atoms on both sides.
\(3Zn + HNO_3
ightarrow3Zn(NO_3)_2 + H_2\)

Step2: Balance nitrogen atoms

In \(3Zn(NO_3)_2\), there are 6 nitrogen atoms. So we need 6 nitrogen atoms on the left - hand side. By putting a coefficient of 6 in front of \(HNO_3\), we have 6 nitrogen atoms on both sides.
\(3Zn + 6HNO_3
ightarrow3Zn(NO_3)_2 + H_2\)

Step3: Balance hydrogen and oxygen atoms

On the left - hand side, we have 6 hydrogen atoms in \(6HNO_3\). On the right - hand side, in \(3Zn(NO_3)_2\) there are no hydrogen atoms. In \(H_2\), we need to adjust the coefficient to balance hydrogen. Also, for oxygen atoms, after adjusting the coefficients, we find that the correct balanced equation for the reaction of zinc with nitric acid to produce zinc nitrate and hydrogen is:
\(3Zn + 8HNO_3
ightarrow3Zn(NO_3)_2+4H_2O + 2NO\) (Note: the original reaction as written is incorrect as nitric acid does not react with zinc in this simple way to produce hydrogen directly. The correct reaction of zinc with dilute nitric acid is a redox reaction. But if we assume a non - redox acid - metal reaction for the sake of simple balancing of the given skeletal equation, we have:
For the given equation \(3Zn + 2HNO_3
ightarrow3Zn(NO_3)_2 + H_2\), to balance hydrogen and oxygen, we rewrite it as \(3Zn+8HNO_3 = 3Zn(NO_3)_2+4H_2O + 2NO\). But if we force the given form to balance just the elements shown:
We need 8 moles of \(HNO_3\) to balance hydrogen and oxygen properly for the given products. And the coefficient of \(H_2\) should be 4.

Answer:

4