posted by Luke on .
A mixture of ammonia and oxygen is prepared by combining 0.250 L of NH3 (measured at 0.700 atm and 20°C) with 0.160 L of O2 (measured at 0.720 atm and 49°C). How many milliliters of N2 (measured at 0.740 atm and 100. °C) could be formed if the following reaction occurs?
4NH3(g) + 3O2(g) -->2N2(g) + 6H2O(g)
I get that oxygen is the limiting factor
For NH3 I get n=.00727
For o2 I get n= .00435
So I do 3/2 x .00435 =.0029
Then .740atm x L = .0029 x .0821 x 373
Then .0888/.740 =.120
Does this look right?
Some picky points.
n N2 I obtained 0.007279 so I rounded to 0.00728.
For n O2 I obtained 0.00436
I obtained 0.0029 for mols N2 just as you did; however, it is done by 0.0036 x 2/3 and not 3/2.
Finally,(this is not so picky) the problem asks for mL so you should convert your 0.120L to 120 mL.
Thank you! It was right!!!