Answer:
0.576 mol of KMnO₄
91.0 g of KMnO₄
Explanation: -
Molar mass of KNO₃ = 39 x 1 + 14 x 1 + 16 x 3
= 101 g/ mol
Mass of KNO₃ =145 grams
Number of moles of KNO₃ =145 g / (101 g/ mol)
= 1.44 mol
The chemical equation for the reaction is
5 KNO₂ + 2KMnO₄ + 3 H₂SO₄ →5 KNO₃ + K₂SO₄ + MnSO₄ + 3 H₂O
From the balanced chemical equation we see
5 mol of KNO₃ is produced from 2 mol of KMnO₄
1.44 mol of KNO₃ is produced from [tex] \frac{2 mol KMnO4 x 1.44 mol KNO3 }{5 mol KNO3} [/tex] mol of KMnO₄
= 0.576 mol of KMnO₄
Molar mass of KMnO₄ = 39 x 1 + 55 x 1 + 16 x 4
= 158 g /mol
Mass of KMnO₄ = 158 g /mol x 0.576 mol of KMnO₄
= 91.0 g
Answer:
0.5742 moles and grams of potassium manganate are needed to provide 145 grams of potassium nitrate.
Explanation:
Mass of potassium nitrate needed = 145 g
Moles of potassium nitrate = [tex]\frac{145 g}{101 g/mol}=1.4356 mol[/tex]
[tex]5KMnO_2+2KMnO_4+3H_2SO_4\rightarrow 5KNO_3+2MnSO_4+K_2SO_4+3H_2O[/tex]
According to reaction, 5 moles of potassium nitrate are obtained from 2 moles of potassium manganate .
Then 1.4356 moles of potassium nitrate will be obtained from:
[tex]\frac{2}{5}\times 1.4356 mol=0.5742 mol[/tex]
Mass of 0.5742 moles of potassium manganate :
0.5742 mol × 158 g/mol = 90.72 g
0.5742 moles and grams of potassium manganate are needed to provide 145 grams of potassium nitrate.
