Respuesta :
Answer: a) Volume of [tex]O_2][/tex] = 19.2 L
b) volume of [tex]CO_2[/tex] = 20.8 Liters
Explanation:
Combustion is a type of chemical reaction in which fuel is reacted with oxygen to form carbon dioxide and water.
[tex]C_6H_{12}O_6+6O_2 \rightarrow 6CO_2+6H_2O[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{23.5g}{180g/mol}=0.131moles[/tex]
a) Volume of [tex]O_2[/tex]
According to stoichiometry:
1 mole of glucose require = 6 moles of oxygen
Thus 0.131 moles of glucose require =[tex]\frac{6}{1}\times 0.131=0.783[/tex] moles of oxygen
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1.00 atm
V= Volume of the gas = ?
T= Temperature of the gas = 298 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.783
[tex]V=\frac{nRT}{P}=\frac{0.783\times 0.0821\times 298}{1.00}=19.2L[/tex]
Thus volume of oxygen required is 19.2 Liters
b) Volume of [tex]CO_2[/tex]
According to stoichiometry:
1 mole of glucose produce = 6 moles of [tex]CO_2[/tex]
Thus 0.131 moles of glucose require =[tex]\frac{6}{1}\times 0.131=0.783[/tex] moles of [tex]CO_2[/tex]
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 0.960 atm
V= Volume of the gas = ?
T= Temperature of the gas = [tex]37^0C[/tex]=310 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.783
[tex]V=\frac{nRT}{P}=\frac{0.783\times 0.0821\times 310}{0.960}=20.8L[/tex]
Thus volume of [tex]CO_2[/tex] produced is 20.8 Liters
