Question

a mixture of CH4,N2 and O2 is enclosed in a container of 1 litre capacity at 0 degree celsius total pressure of gaseous mixture is 2660 mm Hg.if the ratio of partial pressure of the gases is 1:4:2 respectively,the number of moles of oxygen present in the vessel is

Answer 1

the formula of Ch 4 is methane the formula of N2 is nitrogen its chemical name I say and O2 is Oxygen gas ok now the number of moles of oxygen present in the vessel is 2

Answer 2

Answer:

The number of moles of oxygen present in the vessel is 0.0446 moles.

Explanation:

Total pressure of the mixture in container,P = 2660 mm Hg

The ratio of partial pressure of the gases is 1:4:2.

Let the partial pressure of methane be $p_1=1x$

Let the partial pressure of nitrogen be $p_2=4x$

Let the partial pressure of oxygen be $p_3=2x$

Moles of methane in the mixture $n_1$

Moles of nitrogen in the mixture $n_2$

Moles of oxygen in the mixture $n_3$

According Dalton law of partial pressure:

$P=p_1_p_2+p_3$

$2660 mm Hg=1x+4x+2x$

x = 380 mmHg

Total  moles of gases in mixture =n'

$n'=n_1+n_2+n_3$

PV = nRT

where,

P = Pressure of mixture = 2660 mmHg

V = Volume of carbon monoxide = 1L

n = Number of moles of

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of mixture= 0°C = 273.15 K

$n'=\frac{2660 mmhg \times 1L}{62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 273.15 K}$

n' = 0.1561 moles

$p_1=P\times \chi_1=P\times \frac{n_1}{n'}$

$380 mmHg=2660 mmHg\times \frac{n_1}{0.1561 mol}$

$n_1=0.0223 mol$

$p_2=P\times \chi_2=P\times \frac{n_2}{n'}$

$4\times 380 mmHg=2660 mmHg\times \frac{n_2}{0.1561 mol}$

$n_2=0.0892 mol$

$p_3=P\times \chi_3=P\times \frac{n_3}{n'}$

$2\times 380 mmHg=2660 mmHg\times \frac{n_3}{0.1561 mol}$

$n_3=0.0446 mol$