Question

Using the formula 5KMnO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. How many moles and grams of KMnO4 are needed tk provide 145 grams of KNO3?

Answer 1

0.5742 moles and grams of potassium manganate are needed to provide 145 grams of potassium nitrate.

Step-by-step explanation:

Mass of potassium nitrate needed = 145 g

Moles of potassium nitrate =
`(145 g)/(101 g/mol)=1.4356 mol`

`5KMnO_2+2KMnO_4+3H_2SO_4\rightarrow 5KNO_3+2MnSO_4+K_2SO_4+3H_2O`

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:

`(2)/(5)* 1.4356 mol=0.5742 mol`

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.

Answer 2

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
`(2 mol KMnO4 x 1.44 mol KNO3 )/(5 mol KNO3)` 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