# How many moles of ethylene (C_2H_4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin?

Jul 9, 2016

$0.212 \setminus m o l . {C}_{2} {H}_{4}$

#### Explanation:

Start by using the ideal gas equation to figure out the the number of moles ${O}_{2}$ used.

$P \cdot V = n \cdot R \cdot T$

Where

$P \text{ }$ is the pressure expressed in atm.

$V \text{ }$ is the volume expressed in L.

$n \text{ }$ is the number of moles.

$R \text{ }$ is the universal gas constant, it has a value of $0.0821 \setminus L \cdot a t m . m o {l}^{-} 1 \cdot {K}^{-} 1$

$T \text{ }$is the Kelvin temperature.

Now rearrange the ideal gas equation and solve for n.

$n = \frac{P \cdot V}{R \cdot T}$

$n = \frac{12.9 \setminus L \times 1.2 \setminus a t m}{0.0821 \setminus L \cdot a t m . m o {l}^{-} 1 \cdot {K}^{-} 1 \times 297 \setminus K}$

$n = \frac{12.9 \setminus \cancel{L} \times 1.2 \setminus \cancel{a t m}}{0.0821 \setminus \cancel{L} \cdot \cancel{a t m} \cdot m o {l}^{-} 1 \cdot \cancel{{K}^{-} 1} \times 297 \setminus \cancel{K}}$

$n = 0.635 \setminus m o l .$

In the second step, write a balanced chemical equation for the reaction and use the stoichiometry of the equation to figure out the moles of ethylene used.

${C}_{2} {H}_{4} + 3 {O}_{2} \to 2 C {O}_{2} + 2 {H}_{2} O$

$0.635 \setminus m o l . {O}_{2} \times \frac{1 \setminus m o l . \setminus {C}_{2} {H}_{4}}{3 \setminus m o l . \setminus {O}_{2}}$

$0.635 \setminus \cancel{m o l . {O}_{2}} \times \frac{1 \setminus m o l . \setminus {C}_{2} {H}_{4}}{3 \setminus \cancel{m o l . \setminus {O}_{2}}}$

$0.212 \setminus m o l . {C}_{2} {H}_{4}$