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Answers
The first reaction is NH3 --> 0.5 N2 + 1.5 H2
The second one is H2 + 0.5 O2 --> H2O
10 ml NH3 react to 0.5 * 10 ml N2 and 1.5 *10 ml H2. Total volume is 20 ml.
The 15 ml H2 react with halve the volume of O2 (7.5 mL) to give 15 ml of H2O(g).
So the overall calculus of volumes is:
5 ml N2, 15 ml of H2O(g), 30 - 7,5 = 22,5 mL of O2.
This is only true if we assume that the water is liquid and its volume can be neglected.
I don't know what we need the vapor density of ammonia and the oxidation to H2O for. The volume change of the first equation (10 mL to 20 mL) only works, if the ration of N to H is 1/3.
Explanation:
If assumed we are dealing with ideal gases (thats not true) we know that 1 mole of that gas has a volume (at STP) of 22.4 L. So the gas volumes can be set to moles.
The first reaction is NH3 --> 0.5 N2 + 1.5 H2
The second one is H2 + 0.5 O2 --> H2O
10 ml NH3 react to 0.5 * 10 ml N2 and 1.5 *10 ml H2. Total volume is 20 ml.
The 15 ml H2 react with halve the volume of O2 (7.5 mL) to give 15 ml of H2O(g).
So the overall calculus of volumes is:
5 ml N2, 15 ml of H2O(g), 30 - 7,5 = 22,5 mL of O2.
This is only true if we assume that the water is liquid and its volume can be neglected.
I don't know what we need the vapor density of ammonia and the oxidation to H2O for. The volume change of the first equation (10 mL to 20 mL) only works, if the ration of N to H is 1/3.