Question

0.1 mole of potassium permanganate was heated at 300°C. What is the weight (in g) of the residue?

• A. 14.2 g
• B. 1.6 g
• C. 15.8 g
• D. 7.1 g

Hint:

In reactions involving thermal decomposition, calculate the amount of residue using the stoichiometry of the reaction and the molar masses of the compounds involved.

Answer 1

The Correct Option is A

To determine the weight of the residue upon heating potassium permanganate (KMnO₄) at 300°C, we need to understand the decomposition reaction:
2 KMnO₄ → K₂MnO₄ + MnO₂ + O₂↑

This reaction shows that 2 moles of KMnO₄ decompose to produce 1 mole of K₂MnO₄ and 1 mole of MnO₂, releasing oxygen gas.

The molar masses are:

• KMnO₄: 158 g/mol
• K₂MnO₄: 197 g/mol
• MnO₂: 87 g/mol

From the balanced equation, decomposing 2 moles of KMnO₄ (2 x 158 g = 316 g) produces 1 mole of K₂MnO₄ (197 g) and 1 mole of MnO₂ (87 g). Therefore, the total residue is:

K₂MnO₄ + MnO₂ = 197 g + 87 g = 284 g

The reaction ratio is 316 g of KMnO₄ giving 284 g of residue. Now, calculate for 0.1 mole of KMnO₄:

0.1 mole of KMnO₄ = 0.1 x 158 g = 15.8 g

Using the ratio:

(284 g residue / 316 g KMnO₄) x 15.8 g KMnO₄ = x g residue

Solving for x:

x = (284/316) x 15.8 ≈ 14.2 g

Therefore, the weight of the residue is 14.2 g.

Answer 2

We are given that 0.1 mole of potassium permanganate is heated at 300°C. The molar mass of potassium permanganate (KMnO₄) is 158 g/mol. The residue is the mass of potassium manganate (K₂MnO₄).
From the reaction:

\( 2 KMnO_4 \xrightarrow{{heat}} K_2MnO_4 + O_2 \)

For 2 moles of KMnO₄, 1 mole of K₂MnO₄ is produced. This means the ratio of KMnO₄ to K₂MnO₄ is 2:1.
The molar mass of K₂MnO₄ is calculated as:
 

\( \text{Molar mass of K}_2{MnO}_4 = 2(39) + 55 + 4(16) = 158 \, {g/mol} \)

Thus, the mass of K₂MnO₄ formed from 0.1 mole of KMnO₄ is:

\( \frac{158}{2} \times 0.1 = 14.2 \, {g} \)

Thus, the weight of the residue is 14.2 g.