Part a three gases 800 g of methane ch4 180 g of ethane c2h6...

Part a. Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 °C, the total pressure in the container is 4.90 atm. Calculate the partial pressure of each gas in the container. Part b. A gaseous mixture of O2 and N2 contains 38.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 245 mmHg?

Transcript

00:01 Hello students, here in this question we have given these information we have to first calculate the number of moles.

00:06 We know that pv equals to nrt or we can write this as n is equals to pv by rt.

00:13 So when we put the values 5 .7 atm into 10 liters and r is the gas constant that is 0 .0821 here and the temperature given 296 kelvin.

00:31 Now on solving we will get number of moles of total solution that is 2 .18 mole.

00:38 Now we have to find out the number of moles of each gas.

00:42 So the number of moles of methane will be equal to the mass over molecular weight.

00:51 So the mass we have given in question that is 8 gram and the molecular weight that is 16.

01:00 So we will get this as 0 .499 mole.

01:07 From here gram will be cancelled out.

01:09 Now the number of moles of ethane, ethane it will be same mass over molecular weight.

01:23 Mass we have given that is 18 gram and the molecular weight is 30 gram per mole.

01:30 From here gram gram will cancel out we will get the number of moles 0 .599.

01:37 Now we have to calculate the moles of propane.

01:39 So we know that total moles.

01:43 So from here we can calculate the number of moles of propane.

01:45 The total moles here is 2 .18.

01:49 The moles of methane that is 0 .499 and for methane it is 0 .599 and we have to calculate the number of moles of propane.

02:00 So it will be equal to 1 .08...

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