Chemistry Daily Challenge 22-July-2015

Chemistry Level 2

What is the ratio of the coefficients of $\ce{MnO4-}$ , $\text{C}_2\text{O}_4^{2-}$ , and $\ce{H+}$ when the following reaction equation is balanced?

$\ce{MnO4-}+\text{C}_2\text{O}_4^{2-}+\ce{H+} \to \ce{Mn}^{2+}+\ce{CO2}+\ce{H2O}$

2:5:16 2:16:5 5:16:2 16:5:2

1 solution

Chew-Seong Cheong
Jul 24, 2015

Let the coefficients be:

$aMnO_4^- + bC_2O_4^{2-} + cH^+ \rightarrow aMn^{2+} + 2bCO_2 + \dfrac{c}{2}H_2O$

Equating for the number of $O$ , we have:

$4a + 4b = 4b + \dfrac{c}{2} \quad \Rightarrow c = 8a$

The only answer option acceptable is $\boxed{2,5,16}$ . Substituting the coefficients in the reaction equation we get a balanced one:

$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O \\ 2Mn, (8+20)O, 10C, 16H = 2Mn, (20+8)O, 10C, 15H$

Moderator note:

By right, we have 6 variables to start with. However, by comparing the elements of $\ce{Mn}$ , $\ce{C}$ and $\ce{H}$ , we can reduce it by 3 variable hence resulting in just $a , b, c$ .

Elegant solution......sir!!!!!!

Aamir Faisal Ansari - 5 years, 10 months ago

Chemistry Daily Challenge 22-July-2015

1 solution

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