what industrial preparation of trioxocarbonate acid ammonia is burn in oxygen in the presence of a catalyst according to th following equation 4NH3(g)+5O2(g)==>4NO3+6H2O(g) If 250cm3 are burnt completely what valume of oxygen is used up
I'm sorry but I don't get it. What in the world is trioxocarbonate acid ammonia. The equation can't be right. Never heard of NO3. It isn't balanced either. Are you sure you aren't trying to write the equation for the industrial preparation of HNO3 (not NO3) from the oxidation of NH3. That's the Haber process.
Think moles.
5/9 of the reactants are O2, so I'd say 5/9 * 250 cm^3 were O2
160
There is no solving
To determine the volume of oxygen used up in the reaction, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction is:
4NH₃(g) + 5O₂(g) → 4NO₃(g) + 6H₂O(g)
From the equation, we can see that 4 moles of NH₃ react with 5 moles of O₂. This means that the ratio of NH₃ to O₂ is 4:5.
Since we are given the initial volume in cm³, we need to convert it to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed constant)
V = volume (cm³)
n = moles
R = ideal gas constant
T = temperature (assumed constant)
Given that the pressure, temperature, and volume are constant for both NH₃ and O₂, the ratio of their moles can be determined directly from their volumes.
If we let x be the volume of O₂ used up (in cm³), then based on the 4:5 ratio, the volume of NH₃ used up would be (4/5) * x.
Since the moles of gas are directly proportional to their volume, we can say:
(4/5) * x (cm³) of NH₃ reacts with x(cm³) of O₂.
Therefore, if 250 cm³ of NH₃ are burned completely, we can set up the following equation:
(4/5) * x = 250
Now we can solve for x:
x = (250 * 5) / 4
x = 312.5 cm³
So, 312.5 cm³ of oxygen would be used up in the reaction.