When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bold type? MnO₂ + HCl$\rightarrow$ MnCl₂ + Cl₂ + H₂O A) 5 B) 4 C) 3 D) 2 E) 1

Question

Quizplus · Accepted Answer

First, we balance the sulfur by adding a coefficient of 6 in front of HCl:

MnSO4 + 6HCl → MnCl2 + 2Cl2 + 3H2O + 2HSO4

Now we balance the chlorine by adding a coefficient of 4 in front of MnSO4:

4MnSO4 + 24HCl → 4MnCl2 + 8Cl2 + 12H2O + 8HSO4

Finally, we balance the manganese by adding a coefficient of 5 in front of HCl:

4MnSO4 + 120HCl → 4MnCl2 + 8Cl2 + 12H2O + 8HSO4

So the number in front of MnSO4 is 4, and the answer is B.

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When the Following Equation Is Balanced Using the Smallest Possible $\rightarrow$