Question

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Answer 1

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

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.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$$n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$$n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

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.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$$V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$$V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

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