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.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Lat STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

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

$\text{Number of moles of propane}=\frac{\text{Given volume}}{\text{Molar volume}}=\frac{15.0L}{22.4L}=0.67moles$

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

According to stoichiometry:

1 mole of propane combines with = 5 moles of oxygen

Thus 0.67 moles of propane combine with  = $\frac{5}{1}\times 0.67=3.35moles$

Volume of $O_2=moles\times {\text {Molar Volume}}=3.35\times 22.4L=75L$

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.