Question

What volume (in L) of water vapor will be

produced from the reaction of 24.65 L of oxygen?

2C2H6(g) + 702(g) —4CO2(g) + 6H2O(9)

Answer 1

21.13

Step-by-step explanation:

Answer 2

Answer: 21.13 L of
`H_2O` will be produced from the reaction of 24.65 L of oxygen

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

`2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)`

According to stoichiometry :

7 moles of
`O_2` give = 6 moles of
`H_2O` vapour

`7* 22.4L=156.8L` of
`O_2` produce =
`6* 22.4L=134.4L` of
`H_2O` vapour

Thus 24.65 L of
`O_2` produce =
`(134.4)/(156.8)* 24.65=21.13` L of
`H_2O` vapour

21.13 L of
`H_2O` will be produced from the reaction of 24.65 L of oxygen