# How do you calculate the partial pressures of this problem?

## A mixture containing 2.53 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50 L flask at a temperature of 25°C. (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture. Answers in atmospheres.

Aug 13, 2016

In a gaseous mixture, the pressure exerted by a component is the same as if it ALONE occupied the container............

#### Explanation:

.........and the total pressure is the sum of the individual partial pressures.

The above was a restatement of Dalton's Law of Partial Pressures, which was a very early experimental finding. From this, to find the pressure exerted by a component, all we need to do is find the total number of moles of each gas, and proceed from there.

In your problem there are $\frac{2.53 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} \text{ methane; "(2.53*g)/(28.05*g*mol^-1)" ethylene"; (2.53*g)/(58.12*g*mol^-1)" butane} .$

Now we may use Dalton's Law to get the individual partial pressures:

P_"methane"=(n_"methane"RT)/V $=$ $\frac{2.53 \cdot g \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 298 \cdot K}{16.04 \cdot g \cdot m o {l}^{-} 1 \times 1.50 \cdot L}$ $=$ ??atm

And ${P}_{\text{Total"=P_"methane"+P_"ethylene"+P_"butane}}$

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