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In the reaction, VO32– + MnO4–1 → Mn2+ + VO43–, 1 mole of VO32– is oxidized by how many moles of MnO4–? [rounded up to second decimal place]
Correct answer is '0.20'. Can you explain this answer?
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Inthe reaction, VO32– + MnO4–1 → Mn2+ + VO43–, ...
Explanation:
Step 1: Determine the balanced equation
The balanced equation for the reaction is:
2 VO32– + 5 MnO4– + 6 H+ → 5 Mn2+ + 2 VO43– + 3 H2O
Step 2: Calculate the mole ratio
From the balanced equation, we can see that 2 moles of VO32– are oxidized by 5 moles of MnO4–.
Step 3: Calculate the mole ratio between VO32– and MnO4–
To find the mole ratio between VO32– and MnO4–, we divide the coefficients of VO32– and MnO4– in the balanced equation:
2 moles VO32– / 5 moles MnO4– = 0.4 moles VO32– / 1 mole MnO4–
Step 4: Round up to the second decimal place
The mole ratio between 1 mole of VO32– and MnO4– is approximately 0.40. When rounded up to the second decimal place, it becomes 0.40, which is equivalent to 0.20.
Therefore, 1 mole of VO32– is oxidized by approximately 0.20 moles of MnO4–.
Step 1: Determine the balanced equation
The balanced equation for the reaction is:
2 VO32– + 5 MnO4– + 6 H+ → 5 Mn2+ + 2 VO43– + 3 H2O
Step 2: Calculate the mole ratio
From the balanced equation, we can see that 2 moles of VO32– are oxidized by 5 moles of MnO4–.
Step 3: Calculate the mole ratio between VO32– and MnO4–
To find the mole ratio between VO32– and MnO4–, we divide the coefficients of VO32– and MnO4– in the balanced equation:
2 moles VO32– / 5 moles MnO4– = 0.4 moles VO32– / 1 mole MnO4–
Step 4: Round up to the second decimal place
The mole ratio between 1 mole of VO32– and MnO4– is approximately 0.40. When rounded up to the second decimal place, it becomes 0.40, which is equivalent to 0.20.
Therefore, 1 mole of VO32– is oxidized by approximately 0.20 moles of MnO4–.
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