Questions: HNO3 + S → H2SO4 + NO You now have these half-reaction equations: 3 e^- + HNO3 → NO S → H2SO4 + 6 e^- Which factor will you use for the top equation? Which factor will you use for the bottom equation?

HNO3 + S → H2SO4 + NO

You now have these half-reaction equations:
3 e^- + HNO3 → NO
S → H2SO4 + 6 e^-

Which factor will you use for the top equation? 
Which factor will you use for the bottom equation?
Transcript text: \[ \mathrm{HNO}_{3}+\mathrm{S} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO} \] You now have these half-reaction equations: \[ \begin{array}{l} 3 \mathrm{e}^{-}+\mathrm{HNO}_{3} \rightarrow \mathrm{NO} \\ \mathrm{~S} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+6 \mathrm{e}^{-} \end{array} \] Which factor will you use for the top equation? $\square$ Which factor will you use for the bottom equation? $\square$
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Solution

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Solution Steps

Step 1: Determine the Least Common Multiple (LCM) of Electrons
  • The top half-reaction involves 3 electrons.
  • The bottom half-reaction involves 6 electrons.
  • The LCM of 3 and 6 is 6.
Step 2: Calculate the Multiplication Factor for Each Half-Reaction
  • For the top half-reaction: \( \frac{6}{3} = 2 \)
  • For the bottom half-reaction: \( \frac{6}{6} = 1 \)
Step 3: Apply the Multiplication Factors
  • Multiply the top half-reaction by 2.
  • Multiply the bottom half-reaction by 1.
Step 4: Output the Factors
  • Factor for the top equation: \( 2 \)
  • Factor for the bottom equation: \( 1 \)

Final Answer

The factor for the top equation is \( 2 \) and the factor for the bottom equation is \( 1 \). Thus, the answers are:

  • Top equation factor: \( \boxed{2} \)
  • Bottom equation factor: \( \boxed{1} \)
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