Question

What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?

2c2H6(g)+7O2(g)--->4CO2(g)+6H2O(g)

Answer 1

90.3

Step-by-step explanation:

Answer 2

For the production of 77.4 L water 90.3 L oxygen is required.

Step-by-step explanation:

Given data:

Volume of oxygen required = ?

Volume of water produced = 77.4 L

Solution:

Chemical reaction equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

1 mole = 22.414 L

There are 6 moles of water = 6×22.414 = 134.5 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

Now we will compare the litters of water and oxygen:

H₂O : O₂

134.5 : 156.9

77.4 : 156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.