Question

Saved Propane burns in air according to the equation C3Ha(g 502lg)3CO2) + 4H20(g) What volume of O2 in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions? Short Answer Toolbar navigation E I E B IUS EA This question will be sent to your Instructor for grading. 20 of 25 l Next > Prev nere to search

Answer 1

Answer: 75 liters of
`O_2` in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.

Step-by-step explanation:

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

To calculate the number of moles, we use the equation:

`\text{Number of moles of propane}=\frac{\text{Given volume}}{\text{Molar volume}}=(15.0L)/(22.4L)=0.67moles`

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

According to stoichiometry:

1 mole of propane combines with = 5 moles of oxygen

Thus 0.67 moles of propane combine with =
`(5)/(1)* 0.67=3.35moles`

Volume of
`O_2=moles* {\text {Molar Volume}}=3.35* 22.4L=75L`

Thus 75 liters of
`O_2` in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.