Avolume of 18.0 l contains a mixture of 0.250 mole n2 , 0.250 mole o2 , and an unknown quantity of he . the temperature of the mixture is 0 ∘c , and the total pressure is 1.00 atm . how many grams of helium are present in the gas mixture?

There are 1.22 grams of He in the mixture. The mixture contains 0.304 moles of He, 0.250 moles of N2 and 0.250 moles  of O2. Since the mixture of gases are at standard temperature and pressure, the total moles of the mixture equivalent to 18.0 L is 0.804 moles.

Further Explanation:

At standard temperature and pressure of 0°C and 1.00 atm, the molar volume of gas is 22.4 L.

From this, the total moles of the mixture can be derived using the solution below:

Therefore, the total moles of gases in an 18.0 L mixture at STP is:

By algebra, the moles of He may be solved using the given moles of the other two gases: 0.250 mol N2 and 0.250 mol O2.

0.304 moles of He is equal to 1.22 grams of He.

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Keywords: molar volume, STP, stoichiometry

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.