How to find number of moles of HNO3 behaving like acid in a redox reaction?

Question

In the reaction $(1)$, how to find the number of moles of $\ce{HNO3}$, which is behaving like acid?

The reaction is given as follows:

$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O} \tag1$$

To find number of moles of $\ce{HNO3}$ behaving like acid, I first balanced the reaction using oxidation number method which gave me:

$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

But I could not proceed any further. I am unable to find any hint. How should we proceed to reach the answer? Answer mentioned in book is 9. This is a question from chapter "Redox Reactions".

Answer 1

It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:

$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$

Thus, the two relevant half reactions to this redox equation are: $$\ce{Sn (s) -> Sn^2+ (aq)} \\ \ce{NO3- (aq) -> NH4+ (aq)}$$

First balance elements other than $\ce{O}$ and $\ce{H}$. In this case they are $\ce{Sn}$ in first equation and $\ce{N}$ in second equation. Then, balance $\ce{O}$ and $\ce{H}$ in both equations with water and $\ce{H+}$, respectively. This is because the reaction is in aqueous acidic medium:

$$\ce{ Sn (s) -> Sn^2+ (aq)}\\ \ce{NO3- (aq) + 10 H+ (aq) -> NH4+ (aq) + 3 H2O (l)}$$

Finally, balance the positive charges by electrons so that each equation has no net charges: $$\ce{ Sn (s) -> Sn^2+ (aq) + 2e-} \tag1$$ $$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -> NH4+ (aq) + 3 H2O (l)} \tag2$$

These are your mass and charge balanced oxidation ($(1)$) and reduction ($(2)$) half-reactions. Now, you can add these two equations in order to cancel electrons. To do so add $4 \times (1) + (2)$:

$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -> 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$

This is your balanced ionic equation. As evidence, you need nine $\ce{NO3-}$ ions to balance as counter ions in right hand side, so you need to add $\ce{9NO3-}$ to both side to keep the mass and charge balance.:

$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -> 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$

When simplify, the equation $(4)$ would look like:

$$\ce{ 4 Sn +10 H+ + 10 NO3- -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$

Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:

$$\ce{ 4 Sn +10 HNO3 -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$

Therefore, your answer is correct. Answer in your textbook must be a misprint or accidently for got that one $\ce{HNO3}$ is needed to be reduced. Other $\ce{9NO3-}$ is to balance the products (as counter ions). All $\ce{10H+}$ ions needed to complete the redox reaction.

Answer 2

While balancing, $$\ce{Sn + HNO_{3} \longrightarrow Sn(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O}$$

first balance oxidising agent and reducing agent only, we will get $\ce{4Sn}$ and $\ce{1HNO_{3}}$ , after balancing them, write number of extra moles of acid obtained while balancing other atoms. That will give us the answer.