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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?

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    $\begingroup$ So for the nonsensical reaction what is your balanced chemical formula? In other words the Zn, H, N, and O atoms have to balance on both sides of the reaction. // Obviously $x = a$, so $$\ce{aZn + yHNO3 -> aZn(NO3)2 + bH2O + cNH4NO3}$$ $\endgroup$ – MaxW Apr 27 at 18:45
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    $\begingroup$ It's 4 ,10,4,1,3 . I was wrong earlier . The answer is 4+3+1 which is eight $\endgroup$ – user198885 Apr 27 at 18:56
  • $\begingroup$ Curious, did you pound your head on the wall guessing, or did you use the set of linear equations for the atoms? $\endgroup$ – MaxW Apr 27 at 19:02
  • $\begingroup$ Linear equations $\endgroup$ – user198885 Apr 27 at 19:19
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    $\begingroup$ I sincerely don't understand why the question is downvoted. The question itself looks reasonable, the reaction is chemically correct, and OP demonstrated their efforts and thoughts (partially in the comments, but still). $\endgroup$ – andselisk Apr 27 at 22:17
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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 $a + b + c = 4 + 3 + 1 = 8$.

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    $\begingroup$ I don't understand why it was commented that this is a nonsensical reaction? Nitrate ion can be easily reduced to ammonium ions. $\endgroup$ – M. Farooq Apr 28 at 5:36
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    $\begingroup$ @M.Farooq You should ask the one who commented. As I mentioned in the comments and in answer itself the reaction looks completely fine to me. $\endgroup$ – andselisk Apr 28 at 5:37
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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 $c=1$. $$\therefore a+b+c=4+3+1=8$$

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