When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bold type? Na₂S₂O₃ + I₂ $\rightarrow$ NaI + Na₂S₄O₆ A) 1 B) 2 C) 3 D) 4 E) 6

Question

Quizplus · Accepted Answer

The substance in bold type is F1F1F1S1, which appears on both sides of the equation. Therefore, we can eliminate it from the equation and focus on balancing the remaining substances. Starting with ₂S₂O₃ and I₂, we can balance the sulfur by placing a coefficient of 2 in front of I₂, giving us 2I₂. This gives us 2₂S₂O₃ + 2I₂. Next, we balance the iodine by placing a coefficient of 5 in front of NaI, giving us 5NaI. This gives us: 2₂S₂O₃ + 2I₂ + 5NaI. Finally, we balance the sodium by placing a coefficient of 10 in front of Na₂S₄O₆, which gives us the balanced equation: 2₂S₂O₃ + 2I₂ + 5NaI → 10Na₂S₄O₆. Thus, the coefficient of Na₂S₄O₆ is 10, and the answer is B.

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

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