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Ammonia NH3 burns in oxygen to produce nitrogen dioxide NO2 and water. If 20.3 litres of ammonia are burned how many litres of oxygen are required?
Wiki User
∙ 14y ago
Use dimensional analysis
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The reaction stoichiometry is given by: 4NH3 + 7O2 -> 4NO2 + 6H2O
Assuming the reaction is performed in the gas phase at room conditions (1 bar, 298 K) the molar volume of the gases is equal to: Vm = V/n = (RxT)/P = (8.314x10^-2 x 298)/1 = 24.8 L/mol (where R = 8.314x10^-2 bar.L/mol.K)
For every 4 moles of NH3 burned 7 moles of O2 are needed. In the performed reaction 20.3 L NH3 are burned which equals 20.3/24.8= 0.82 mol NH3. The required amount of O2 is then 7x0.82/4 = 1.44 mol O2. The required volume is then 1.44 x 24.8 = 35.6 L O2.
Clarification: To make it even simpler, if you have only gases in the equation, you do not have to calculate the number of moles, you can use the ratio of the two gases and the volume of one to find the volume of the other (Gay-Lussac's Law). In the balanced equation, we have the ratio of NH3 to O as 4:7, so for every 4L of NH3, we need 7L of O2. Now we have (20.3L NH3 / 4) x 7, which equals 35.5L O2.
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∙ 14y ago
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