Respuesta :
1) Write the chemical equation.
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]2) List the known and unknown quantities.
Sample: CH4.
Volume: 2.0 L.
Temperature: 30 ºC = 303.15 K.
Pressure: 3.0 atm.
Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).
Moles: unknown.
3) Moles of CH4.
3.1- Set the equation.
[tex]PV=nRT[/tex]3.2- Plug in the known values and solve for n (moles).
[tex](3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)[/tex][tex]n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=[/tex][tex]n=0.24\text{ }mol\text{ }CH_4[/tex]4) Moles of oxygen that reacted.
The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.
[tex]mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2[/tex]5) Volume of oxygen required.
Sample: O2.
Moles: 0.48 mol.
Temperature: 30 ºC = 303.15 K.
Pressure: 3.0 atm.
Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).
Volume: unknown.
5.1- Set the equation.
[tex]PV=nRT[/tex]5.2- Plug in the known values and solve for V (liters).
[tex](3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)[/tex][tex]V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}[/tex][tex]V=3.98\text{ }L[/tex]3.98 L of O2 is required to react with 2.0 L CH4.
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