Balancing the reaction of nitric acid with zinc using oxidation number method

Question

In a previous question, there was confusion relating to what this "oxidation number method" is. So I will illustrate that before my question.

1. Write the skeleton equation representing chemical change.
2. Assign oxidation numbers to the atoms in the equation and then write separate equations for atoms undergoing oxidation and reduction.
3. Find the change in oxidation number in each equation. Make the change in both equations by multiplying with suitable integers. Add both the equations.
4. Complete the balancing by inspection. First balance those that have undergone change in ON and then other atoms except H and O. Finally, balance the Hs, the Os will be automatically balanced.

Example: $$\ce{Cu + HNO3 -> Cu(NO3)2 + NO2 }$$ $$\ce{Cu^0 -> Cu^{+II}(NO3)2} \tag{1}$$ $$\ce{HN^{+V}O3 -> N^{+IV}O2} \tag{2} $$

To make increase and decrease in ON equal, equation 2 is multiplied by $2$. Adding we get,

$$\ce{Cu + 2 HNO3 -> Cu(NO3)2 + 2NO2 + H2O} \implies \ce{Cu + 4 HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O}$$

Question:

Balance $\ce{Zn + HNO3 -> Zn(NO3)2 + H2O + N2O}$

I am trying to use oxidation number method but unable to reach the right answer.

The half equations are:

$\ce{Zn -> Zn^2+(NO3)2 \tag{1}}$

$\ce{2HN^{+V}O3 -> N^{+I}2O \tag{2}}$

Thus we should multiply the first equation by 2 and then add.

$\ce{2Zn + 2HNO3 -> N2O + 2Zn(NO3)2}$

Then I tried balancing the nitrogens by adding one 2HNO3s to LHS but after that the oxygens ust won't balance!

How do I go about balancing this then?

Answer 1

Here is an alternate- Let the final balanced equation be $\ce{yZn + 2xHNO3-> yZn(NO3)2 +xH2O +zN2O}$

Now For balancing nitrogen we must have $2x=2y+2z$ and for balancing oxygen atoms we should have $6x=6y+x+z$ Now we get an equation $x=5z$. Put $z=1$ you get $x=5$ and $y=4$.

Hence the balanced equation is:

$$\ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + 5H2O + N2O}$$