How much air is required to oxidize 80g of sulfur?

Air is a mixture of gases. Its composition is 78% nitrogen, 21% oxygen and 1% other gases. Combustion is the interaction with oxygen.

Given:
m (S) = 80g
V air = x l

Solution.

S + O2 = SO2

1.According to the equation for the combustion of one mole of sulfur, one mole of oxygen is required.
let’s express their quantity, respectively, in grams and liters:
for 32g S you need 22.4 l of O2,
and for 80g S you need xl O2.
Let’s make the proportion:
32/80 = 22.4 / x
x = 56 (l)
2. Oxygen – 21% of the air composition.
56l – 21%
hl – 100%
x = 56 * 100/21 = 266.6 liters.
Answer. 266.6 L