Answer: 14.7 L of oxygen gas at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53g of propane
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]\text{Number of moles of propane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{5.53g}{44.1g/mol}=0.125moles[/tex]
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
According to stoichiometry :
1 mole of propane requires 5 moles of oxygen
Thus 0.125 moles of propane require=[tex]\frac{5}{1}\times 0.125=0.625moles[/tex] moles of oxygen
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1.04 atm
V= Volume of the gas = ?
T= Temperature of the gas = 25°C = 298 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.625
[tex]V=\frac{nRT}{P}=\frac{0.625\times 0.0821\times 298}{1.04}=14.7L[/tex]
