# What pressure will be exerted by each of the gases in the following mixture if the total pressure of the mixture amoujts of 768.8 torr? .500 g H2, .245 g O2 and .335g N2

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We want to find the pressure exerted by each gas in a mixture of gases. That is, we want to look for the partial pressures of each of the gases. In a mixture of ideal gases, the partial pressure is defined to be the total pressure exerted by a gas if it occupied the container on its own at the same conditions.

Dalton's Law of Partial Pressures states that the total pressure of a mixture of non-reactive gases is simply equal to the sum of the partial pressure of each of the gases. [Note: This holds for ideal and real gases. However, the interpretation that the partial pressure is the pressure of a gas that occupies the container alone, only applies to ideal gases].

`P_(T) = sum_(i=1)^n p_i`

The partial pressure of each gas, moreover, is related to the total pressure, via the mole fraction of each gas:

`p_i = X_i P_T`

Hence, we need to calculate the mole fraction of each given gas.

`0.500g H_2 rightarrow (0.500g H_2) * (1 mol H_2)/(2 g H_2) = 0.25 mol H_2`

`0.245 g O_2 rightarrow (0.245 gO_2)* (1 mol O_2)/(32 g O_2) = 0.0077 mol O_2`

`0.335g N_2 rightarrow (0.335gN_2)*(1molN_2)/(28gN_2) = 0.012 mol N_2`

The total number of moles is: `0.25+0.0077+0.012 = 0.2697 mol`

Hence:

`X_(H_2) = 0.25/0.2697 =0.9270`

`X_(O_2) = 0.0077/0.2697 = 0.0286`

`X_(N_2) = 0.012/0.2697 = 0.0445`

Using the formula for partial pressure:

`P_(H_2) = 0.9270*768.8 t o r r = 712.68 t o r r`

`P_(O_2) = 0.286* 768.8 t o r r = 21.99 t o r r`

`P_(N_2) = 0.0445*768.8 t o r r = 34.21 t o r r`

Therefore, the pressure exerted by the gases are 712.68 torr foy hydrogen, 21.99 torr for oxygen, and 34.21 torr for nitrogen.

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