Finding the sum of coefficients of the reaction


> 
> What is the sum of the coefficients $(a + b + c)$ in the following reaction?
> 
> 
> $$\ce{x Zn + y HNO3} -> {a Zn(NO3)2 + b H2O + c NH4NO3}$$
> 
> 
> 


Here if the equation is balanced we get 1 mol of $\ce{Zn}$ and 4 mol of $\ce{HNO3}$. As $x + y = a + b + c$ we get sum of coefficients $a + b + c = 5$. According to me, this is the correct answer, but in the actual answer key the answer is 8.


I may be wrong, but can someone help me clarify the right answer?




Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me):


$$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -> $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$


$$
\begin{align}
\ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\
\ce{\overset{0}{Zn} &→ \overset{+2}{Zn} + 2 e-} &|\cdot 4 \tag{ox}\\
\hline
\ce{4 \overset{0}{Zn} + \overset{+5}{N} &→ 4 \overset{+2}{Zn} + \overset{-3}{N}} \tag{redox}
\end{align}
$$


So that the complete balanced equation is


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


leaving $a = 4$, $b = 3$, $c = 1$ and the sum $a + b + c = 4 + 3 + 1 = 8$.


 3


> 
> What is the sum of the coefficients $(a + b + c)$ in the following reaction?
> 
> 
> $$\ce{x Zn + y HNO3 -> a Zn(NO3)2 + b H2O + c NH4NO3}$$
> 
> 
> 


This is a redox reaction. During OP's comments, it was clear that he/her got the correct answer, but it wasn't clear that OP understand this is a redox reaction. To balance the equation, it may be better work on with two relevant half reactions. It is also easier to work with ionic equations and add counter ions at the last step. First consider reduction half reaction:
$$\ce{NO3- + d e- -> NH4+}$$
Now, balance $\ce{O}$ with $\ce{H2O}$, $\ce{H}$ with $\ce{H+}$ (since this is in acidic medium), and lastly, charge with $\ce{e-}$ (now you see $d=8$).
$$\ce{NO3- + 10 H+ + 8e- -> NH4+ + 3H2O} \:\text{---} (1)$$
Now, consider oxidation half reaction:
$$\ce{Zn -> Zn^2+ + 2 e- } \:\text{---} (2)$$
Now, solve equation (1) and (2) by cancelling the electrons:
$$\ce{4Zn + NO3- + 10 H+ -> 4 Zn^2+ + NH4+ + 3H2O}$$
Now, to balance by counter ion, add $\ce{9NO3-}$ to both side of last equation so it keep balanced:
$$\ce{4Zn + NO3- + 10 H+ + 9NO3- -> 4 Zn^2+ + NH4+ + 3H2O + 9NO3-}$$
Rewrite the equation with tight pair ions:
$$\ce{4Zn + 10 HNO3 -> 4 Zn(NO3)2 + NH4NO3 + 3H2O }$$
Now you see: $x=4; y=10; a=4; b=3;$ and $c=1$.
$$\therefore a+b+c=4+3+1=8$$


 3