Question

A mixture of 0.220 moles co, 0.350 moles h2 and 0.640 moles he has a total pressure of 2.95 atm. what is the partial pressure of h2?

Answer 1

P = 2.95 atm, total pressure
N = 0.220 + 0.350 + 0.640 = 1.21 moles, total moles

According to Dalton's Law, the partial pressure of H₂ is
(0.350/1.21)*(2.95 atm) = 0.8533 atm

Answer: 0.853 atm

Answer 2

Answer :  The partial pressure of $H_2$ is, 0.853 atm

Solution :

First we have to calculate the mole fraction of $H_2$ in the mixture.

Formula used :

$X_{H_2}=\frac{n_{H_2}}{n_{H_2}+n_{CO}+n_{He}}$

Now put all the given values in this formula, we get

$X_{H_2}=\frac{0.35}{0.22+0.35+0.64}=0.289$

Now we have to calculate the partial pressure of $H_2$

Formula used :

$p_{H_2}=X_{H_2}\times P_T$

where,

$P_T$ = total pressure = 2.95 atm

$p_{H_2}$ = partial pressure of nitrogen gas = ?

$X_{H_2}$ = mole fraction of nitrogen gas = 0.289

Now put all the given values in the above formula, we get

$p_{H_2}=0.289\times 2.95atm$

$p_{H_2}=0.853atm$

Therefore, the partial pressure of $H_2$ is, 0.853 atm