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?
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? $\square$
Solution
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: