# CHEMISTRY HELPP!!?

A 226.4-L cylinder contains 65.5% he(g) and 34.5% kr(g) by mass at 27.0°c and 1.40 ATM total pressure. what is the mass of he in this container? The answer should be 50.3 g but Idk how to get there

THANK YOU!

Relevance

n = PV / RT = (1.40 atm) x (226.4 L) /

((0.082057366 L atm/K mol) x (27.0 + 273.15) K) = 12.869 mol total

Take a hypothetical sample of exactly 100 grams of the gaseous mixture:

(65.5 g He) / (4.002602 g He/mol) = 16.364355 mol He

(34.5 g Kr) / (83.798 g Kr/mol) = 0.411704 mol Kr

(16.364355 mol He) / (16.364355 mol + 0.411704 mol) =

0.975459 (the mole fraction of He)

(12.869 mol total) x (0.975459) x (4.002602 g He/mol) = 50.2 g He

• Login to reply the answers
• First calculate total mol of gas in cylinder:

PV =- nRT

1.40 * 226.4 = n * 0.082057*300

n = 1.40 * 226.4 / ( 0.082057*300)

n = 12.875 total mol

Now take 100 g of the gas mixture:

Mass of He = 65.5 g and mass of Kr = 34.5g

Molar mass He = 4.00 g/mol

Mol He = 65.5g / 4.00 g/mol = 16.375 mol

Molar mass Kr = 83.80 g/mol

mol Kr = 34.5 g / 83.80g/mol = 0.412 mol

Total mol in 100 g gas mixture = 16.375 + 0.412 = 16.787 mol

Total mass gas mixture in cylinder = 12.875 mol / 16.787 mol * 100 g = 76.696 g

Mass of He = 65.5 /100 * 76.696g = 50.24 g

• Login to reply the answers