so I might be just way to tired for this, but I somehow have a stupid mistake here when trying to balance an equation.


Here's what I do:

$$A\cdot Zn(s)+B\cdot MnO_2(s)+C\cdot NH_4^{+}(aq)=D\cdot Zn^{2+}(q)+E\cdot Mn_2O_3(s)+F\cdot NH_3(aq)+G\cdot H_2O(l)\tag{2}$$

We get

$Z_n: \quad A=D$

$Mn:\quad B=2E$

$O:\quad 2B=3E+G$

$N:\quad C=F$

$H:\quad 4C=3F+2G$

$z:\quad C=2D$

We see that

$$Z_n \ \& \ N \ \& \ z: \quad A=D=\frac{1}{2}C=\frac{1}{2}F\tag{3}$$ and $$O \ \& \ Mn:\quad 4E=3E+G \quad \Rightarrow \quad E=G\tag{4}$$ so it follows $$Mn:\quad B=2E=2G\tag{5}$$ we also see $$N \ \& \ H: \quad 4F=3F+2G \quad \Rightarrow \quad F=2G\tag{6}$$ so we get the relationship: $$A=D=\frac{1}{2}C=\frac{1}{2}F=G=E=frac{1}{2}B\tag{7}$$ in correct order: $$A=\frac{1}{2}B=\frac{1}{2}C=D=E=\frac{1}{2}F=G\tag{8}$$

Now the actual solution is: $$Zn(s)+2MnO_2(s)+2NH_4^{+}(aq)=Zn^{2+}(q)+Mn_2O_3(s)+2NH_3(aq)+H_2O(l) \tag{9}$$

so I should have gotten:


I don't see what's wrong. :/

  • $\begingroup$ No, your solution is right because $A = 1$ and $B = 2$, not the other way around. However, you tried to solve a seven variable system of equations without the help of linear algebra, which just made the process more error prone. This problem is even easier if you know about oxidation states. $\endgroup$ – Zhe Jan 16 '19 at 20:58
  • $\begingroup$ I could have made n matrix and solved that, sure. But in general it's not worht the effort for my basic chemistry here. I also just noticed that if I choose $A=1$ then $1=0.5B \Rightarrow B=2$. Thanks $\endgroup$ – xotix Jan 16 '19 at 22:00

Your solution is right, and the answer given is the correct answer. What you have written in equation 8 is incorrect. The numbers you have multiplied in each variable should actually be in the denominator, since they are the ratios.

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