Respuesta :
- The number of moles of KMnO₄ needed to carry out the reaction is 0.0536 mole
- The mass of KMnO₄ needed to carry out the reaction is 8.5 g
Stoichiometry
From the question, we are to determine the number of moles and mass of KMnO₄ needed
From the given balanced chemical equation
5KNO₂ + 2KMnO₄ + 3H₂SO₄ → 5KNO₃ + 2MnSO₄ + K₂SO₄ + 3H₂O
This means 5 moles of KNO₂ is required to react completely with 2 moles of KMnO₄
First, we will determine determine the number of moles of KNO₂ present
From the given information,
Mass of KNO₂ = 11.4 g
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of KNO₂ = 85.1 g/mol
Therefore,
Number of moles of KNO₂ = [tex]\frac{11.4}{85.1}[/tex]
Number of moles of KNO₂ = 0.13396 mole
Now,
If 5 moles of KNO₂ is required to react completely with 2 moles of KMnO₄
Then,
0.13396 mole of KNO₂ will react completely with [tex]\frac{0.13396 \times 2}{5}[/tex] mole of KMnO₄
[tex]\frac{0.13396 \times 2}{5}[/tex] = 0.053584 mole ≅ 0.0536 mole
Hence, the number of moles of KMnO₄ needed is 0.0536 mole
For the mass of KMnO₄ needed,
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of KMnO₄ = 158.034 g/mol
Therefore,
Mass of KMnO₄ needed = 0.053584 × 158.034
Mass of KMnO₄ needed = 8.468 g
Mass of KMnO₄ needed ≅ 8.5 g
Hence, the mass of KMnO₄ needed to carry out the reaction is 8.5 g
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