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:
[tex]total \ mol \ of \ gases \ = 18.0 \ L (\frac{1 \ mol \ }{22.4 \ L})\\[/tex]
Therefore, the total moles of gases in an 18.0 L mixture at STP is:
[tex]\boxed {total \ moles \ of \ gases \ = 0.804\ mol}[/tex]
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.
[tex]0.804 \ mol = 0.250 \ mol \ N_{2} \ + 0.250 \ mol \ O_{2} + x \ mol \ He\\\boxed {mol \ He \ = 0.304 \ mol}[/tex]
0.304 moles of He is equal to 1.22 grams of He.
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Keywords: molar volume, STP, stoichiometry
