@johnsonshelby

Question

@johnsonshelby

callisto · Answer

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 · Answer

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 · Answer

It's so late, I think I did it right...

anonymous · Answer

32.99864952 mol NH3

anonymous · Answer

I have to go back to my dorm, I'll be back on in 15 minutes :)

callisto · Answer

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 · Answer

n = no. of mole of N2 / 0.781

anonymous · Answer

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.