Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation. 4NH3 +5O2-4NO+6H2O.If 27 litres of reactants are consumed ,what volume of nitrogen monoxide is produced at the same temperature and pressure?
To determine the volume of nitrogen monoxide produced, we can use the molar ratio between ammonia and nitrogen monoxide from the balanced chemical equation.
From the equation: 4NH3 + 5O2 -> 4NO + 6H2O
The ratio of NH3 to NO is 4:4, or 1:1.
Given that 27 liters of reactants are consumed, we need to find the volume of nitrogen monoxide produced.
Since the ratio between NH3 and NO is 1:1, the volume of nitrogen monoxide produced will be the same as the volume of ammonia consumed.
Therefore, the volume of nitrogen monoxide produced is 27 liters.
To determine the volume of nitrogen monoxide produced, we first need to calculate the stoichiometric ratio of the reactants and products. According to the balanced equation:
4NH3 + 5O2 -> 4NO + 6H2O
We can see that for every 4 moles of NH3, we produce 4 moles of NO.
To calculate the volume of NO produced, we need to use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the temperature and pressure of the reactants and products are the same, we can assume that the volume ratio is equal to the mole ratio. Therefore, the volume of NO produced will also be 4 times the volume of NH3.
Given that 27 liters of reactants are consumed, we can calculate the volume of nitrogen monoxide produced by multiplying 27 liters by the stoichiometric coefficient:
Volume of NO = 4 * 27 liters
= 108 liters
Therefore, 108 liters of nitrogen monoxide are produced at the same temperature and pressure.
NO is 2/5 of the output of 10 moles.
Since the original 9 moles occupied 27 liters, the 10 moles occupy 30 liters.
So, NO occupies 2/5 * 30 = 12 liters.
This assumes that the H2O is also a vapor.
Total volume of reactants will be 9 moles.
Volume of NO is 4.
When 9 moles of reactants give4 volume of NO then 27liter of reactants will produce what volume of NO.
27*4/9
12LITER of NO.