How many grams of oxygen can be produced by heating 15.8 grams of potassium permanganate to decompose it completely? What is the volume of oxygen under standard conditions (density is 1.429 g / L)?

How many grams of oxygen can be produced by heating 15.8 grams of potassium permanganate to decompose it completely? What is the volume of oxygen under standard conditions (density is 1.429 g / L)?

8 G potassium permanganate was heated to decompose it completely, and the mass of oxygen was X
2KMnO4=K2MnO4+MnO2+O2
316 32
15.8g x 316:32=15.8g:x x=1.6g
Under the standard condition (the density is 1.429g/l), the volume of oxygen is 1.6g divided by 1.429g/l = 1.12l

8 grams of oxygen is needed to burn 3 grams of carbon, so how much air is needed under standard conditions? (oxygen density is 1.429 grams per liter under standard conditions)

Oxygen 8g, standard oxygen volume: 8 / 1.429 = 5.6l
(another method is: do not use density value, mole number: 8 / 32 = 0.25, 0.25 * 22.4 = 5.6 L)
The volume fraction of oxygen in air is 21%, and that of nitrogen is 79%
The required air volume under standard condition is: 5.6 / 0.21 = 26.67l