IMAT 2013 Q51 [Balancing Equations]

What value does c need to be so that the following equation can be balanced?

4KMnO_4 + aH_2SO_4+ 5C_2H_5OH β†’ 4MnSO_4 + bK_2SO_4 + 5CH_3CO_2H + cH_2O

A. 21
B. 11
C. 16
D. 17
E. 26

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The easiest step we can begin with is balancing the number of Potassium atoms.

In place of B, we can see our coefficient should be 2.

Our equation therefore now looks like so:

4KMnO_4 + aH_2SO_4+ 5C_2H_5OH β†’ 4MnSO_4 + 2K_2SO_4 + 5CH_3CO_2H + cH_2O

By adding the 2 as the coefficient for K, we’ve now also changed the amount of SO_4. Which means that in place of a, we need to add a 6.

Our formula now looks as so:

4KMnO_4 + 6H_2SO_4+ 5C_2H_5OH β†’ 4MnSO_4 + 2K_2SO_4 + 5CH_3CO_2H + cH_2O

Since we only have the coefficient c left, which is largely dependent on the amounts of hydrogen in the reactant side- we should count the number of Hydrogen accumulated.

12 from 6H_2SO_4 + 30 from 5C_2H_5OH which totals 42 hydrogen atoms.

We must equate this number on the products side, let’s see how many we hydrogen have pre-existing there;

20 from 5CH_3CO_2H

Meaning we have 22 left to account for (42-20=22). Luckily, we can see than c prefaces H_2O, so the coefficient of hydrogen should simply take half the number we have. 22 divided by 2 is 11. Therefore c is 11, and our answer is B.