3 H2 (g) + N2 (g) --> 2 NH3 (g)
How many liters of ammonia gas can be produced from 10.0 L of N2 and
20.0 L of H2? Assume all gases are measured at the same temperature and pressure.
The first step would be to calculate the MR's of each substance so you could get the reacting masses so:
3H2 = (3 x 2 x 1) = 6
N2 = 14 x 2 = 28
2NH3 = (2 x 14) + (2 x 3 x 1) = 34
so in theory you need 6litres of H2 and 28litres of N2 to produce 34litres of NH3.
Is the question asking for two values or are we to assume that one of them is in excess?
The next step would be to convert to the numbers given in the question. For this I am going to assume that the hydrogen is given in excess.
so instead of 28 litres of nitrogen, I now have 10 litres. First consider 1 litre of nitrogen.
28/28 = 1 so all the other numbers need to be divided by 28.
1litre of N2 and 6/28litres of H2 make 34/28litres of NH3.
so 10 litres would be 10 times all of this.
10 litres of N2 and 10 x (6/28) litres of H2 make 10 x (34/28) litres of NH3.
10 x (34/28) = 12.1 litres of NH3
I know this is confusing, I hope this makes sense.posted by Gemma
what if you want to calculate the number of moles ammonia gas from 3.0L of H2 and 3.0L of N2posted by jerome
4 lposted by ida