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schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Callisto
  • Callisto
Ammonia gas is produced by reacting hydrogen gas with nitrogen gas. To produce 562 g of ammonia gas using air as your source of nitrogen gas, what volume of air is required if the air is introduced at 29°C and 745 mmHg? (the mole fraction of nitrogen gas in air is 0.781)
Callisto
  • Callisto
1. Write the chemical equation \[N_2 + 3H_2 \rightarrow 2NH_3\]2. Find the number of mole of ammonia gas produced. Can you do it?
anonymous
  • anonymous
It's so late, I think I did it right...

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anonymous
  • anonymous
32.99864952 mol NH3
anonymous
  • anonymous
I have to go back to my dorm, I'll be back on in 15 minutes :)
Callisto
  • Callisto
No. of mole of NH3 = mass/ molar mass = 562 / (14.0067 +1.00794x3) = 32.9996 mol No. of mole of N2 required = no. of mole of NH3 / 2 = ... pV = nRT V = 99.32518055000001kPa = 99.32518055000001 x 10^3 Pa T = 29 degree Celsius = (29+273) K Solve V.
Callisto
  • Callisto
n = no. of mole of N2 / 0.781
anonymous
  • anonymous
after you found the moles of NH3 i couldn't follow your work, is there any way that you could please clarify :D i would greatly appreciate it.

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