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

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".

• Try splitting into half-equations to find how many moles of HNO3 are accepting electron pairs and acting as Lewis acids. Jul 31 '20 at 3:32
• I also tried Half reaction method, but I couldn't observe the solution. Jul 31 '20 at 3:33
• @Sahil in the future, enclose chemical equations and formulae in $\ce{}$ : it typesets chemistry automatically. Jul 31 '20 at 4:06
• Jul 31 '20 at 4:35
• Welcome to Chemistry SE! When asking problem statements, could you try to conform to this template? You seem to be someone who would stick through SE for an extended period of time and we would be glad to keep it that way. Jul 31 '20 at 14:49

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.

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.