Question

calculate the amount of oxygen required for the combustion of 89.6 liters
of propane​

Answer 1

Answer:

Given that

Mass of propane =8.8 g

Number of moles =

molar mass

mass

=

44

8.8

=0.2 mol

Now,

C

3

H

8

+ 5O

2

⟶ 3CO

2

+ 4H

2

O

1 mole of propane requires 5×{22.4} litre O

2

.

So, 0.2 mole propane requires 5×{22.4}×{0.2} = 22.4 litre O

2