wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

When heated, potassium permanganate decomposes according to the following equation:

2KMnO4K2MnO4+MnO2+O2

Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 liters.)

Open in App
Solution

Molar mass of KMnO4 =158

Molar volume of O2 at room temperature =24 litre

158 g of KMnO4 at room temperature yields 24 litres of O2

15.8 g of KMnO4 at room temp. will yield =242×158×15.8=1.2 litres of O2.


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mole to Mole relation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon