Air is approximately 79% nitrogen and 21% oxygen by volume. How many moles of oxygen are present in 22.0 L of 24 C and a pressure of 53 kPa?

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Air is approximately 79% nitrogen and 21% oxygen by volume. How many moles of oxygen are present in 22.0 L of 24 C and a pressure of 53 kPa?

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Answer 1 · 2017-05-16T00:09:08

According to Dalton's law of partial pressure, the total pressure of the system is

$P_{t} = n_{t} (\frac{R T}{V})$ $P_t = n_t((RT)/V)$

Convert given units into appropriate units for the gas equations:

$24^{o} C = 297 K$ $24 ^oC = 297 K$

and

$53 k P a = 0.52 a t m$ $53 kPa = 0.52 atm$

The total moles in the system:

$n_{t} = \frac{(0.52 a t m) (22.0 L)}{(0.08206 \frac{L - a t m}{m o l - K}) (297 K)} = 0.47 m o l$ $n_t = ((0.52 atm)(22.0 L))/((0.08206 (L-atm)/(mol-K))(297 K)) = 0.47 mol$

So, using mole fractions of each molecule,

$m o l_{N_{2}} = (0.79) (0.47 m o l) = 0.37 m o l_{N_{2}}$ $mol_(N_2) = (0.79)(0.47 mol) = 0.37 mol_(N_2)$

and

$m o l_{O_{2}} = (0.21) (0.47 m o l) = 0.099 m o l_{O_{2}}$ $mol_(O_2) = (0.21)(0.47 mol) = 0.099 mol_(O_2)$

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Air is approximately 79% nitrogen and 21% oxygen by volume. How many moles of oxygen are present in 22.0 L of 24 C and a pressure of 53 kPa?

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