Callisto
  • Callisto
@johnsonshelby
Chemistry
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schrodinger
  • schrodinger
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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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