Respuesta :
Answer: 75 liters of [tex]O_2[/tex] 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 [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles of propane}=\frac{\text{Given volume}}{\text{Molar volume}}=\frac{15.0L}{22.4L}=0.67moles[/tex]
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
According to stoichiometry:
1 mole of propane combines with = 5 moles of oxygen
Thus 0.67 moles of propane combine with = [tex]\frac{5}{1}\times 0.67=3.35moles[/tex]
Volume of [tex]O_2=moles\times {\text {Molar Volume}}=3.35\times 22.4L=75L[/tex]
Thus 75 liters of [tex]O_2[/tex] in liters would be required if 15.0 L of propane burns, assuming that all of the gases are under the same conditions.
