Question

Calculate the volume of oxygen at step required to burn 7.45 litres of ammonia gas

Answer 1

I think u guys will help me to solve it

Answer 2

4NH3 + 3O2 → 2N2 + 6H2O
Looks like 3/4 as much O2 as NH3

Answer 3

4 NH3 + 5 O2 --> 4 NO2 + 6 H2O
at STP (I think you mean) 22.4 liters/mol of any old perfect gas
7.45 liters * 1 mol/22.4liters) = 0.333 orr about 1/3 of a mol of NH3
we need 5 mols of O2 for every 4 of NH3
so
1/3 mol * 5/4 = 0.4167 mols of O2
0.4167 mols * 22.4 liters/mol = 9.33 liters O2

or we could have just said 7.45 * 5/4 liters

Answer 4

I am not sure if you should burn the nitrogen :)

Answer 5

Aaah! I don't think N2 is the product. At temperatures sufficient to make the reaction NH3 + O2 go, it will be high enough to form NO. As soon as NO is formed it will react with more O2 to produce NO2. So I think the logical equation is
4 NH3 + 5 O2 --> 4 NO2 + 6 H2O (but oxygen is not balanced..See corrected equation at the end).
A search on Google lists a litany of products; i.e., N2, N2O, NO & NO2. With normal amounts of O2 I think N2 is not possible. With insufficient amount of O2 the product of N2O is likely. With normal amounts of O2 both NO and NO2 are feasible. Of course with NO2 and over time you will see equilibrium amounts of N2O4 and possibly N2O3 and N2O5 depending upon conditions. With questions in the realm of general chemistry I would go with 4 NH3 + 7 O2 --> 4 NO2 + 6 H2O and for the question it would be 7.45 L NH3 x (7 mols O2/4 mols NH3) = ? L O2 required or
7.45 x 7/4 = ? L.