How to find number of moles of HNO3 behaving like acid in a redox reaction?
It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -&gt; Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -&gt; Sn^2+ (aq)} \\
\ce{NO3- (aq) -&gt; NH4+ (aq)}$$
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:
$$\ce{ Sn (s)  -&gt; Sn^2+ (aq)}\\
\ce{NO3- (aq) + 10 H+ (aq) -&gt; NH4+ (aq) + 3 H2O (l)}$$
Finally, balance the positive charges by electrons so that each equation has no net charges:
$$\ce{ Sn (s)  -&gt; Sn^2+ (aq) + 2e-} \tag1$$
$$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -&gt; NH4+ (aq) + 3 H2O (l)} \tag2$$
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 $4 \times (1) + (2)$:
$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -&gt; 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$
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.:
$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -&gt; 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$
When simplify, the equation $(4)$ would look like:
$$\ce{ 4 Sn +10 H+ + 10 NO3- -&gt; 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$
Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:
$$\ce{ 4 Sn +10 HNO3 -&gt; 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$
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.