What volume of nitrogen reacts with 33.6 litres of oxygen to produce nitrogen dioxide?
The balanced chemical equation for the reaction between nitrogen and oxygen to produce nitrogen dioxide is:
2NO + O2 → 2NO2
This equation shows that 2 moles of nitrogen react with 1 mole of oxygen to produce 2 moles of nitrogen dioxide.
To calculate the volume of nitrogen required to react with 33.6 litres of oxygen, we need to determine the number of moles of oxygen involved in the reaction using the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin.
Assuming that the temperature and pressure of the gases are constant, we can use the equation to calculate the number of moles of oxygen involved in the reaction:
n(O2) = PV/RT
where P = 1 atm, V = 33.6 L, R = 0.082 L·atm/K·mol, and T = 273 K (standard temperature and pressure conditions).
n(O2) = (1 atm) x (33.6 L) / (0.082 L·atm/K·mol x 273 K) = 1.26 moles
According to the balanced chemical equation, 2 moles of nitrogen react with 1 mole of oxygen to produce 2 moles of nitrogen dioxide. Therefore, the number of moles of nitrogen required is half of the number of moles of oxygen:
n(N2) = n(O2)/2 = 1.26/2 = 0.63 moles
Finally, we can use the ideal gas law again to calculate the volume of nitrogen required:
V(N2) = n(N2)RT/P
where R and T are the same as before, and P = 1 atm.
V(N2) = (0.63 mol) x (0.082 L·atm/K·mol) x (273 K) / (1 atm) = 14.8 L
Therefore, 14.8 litres of nitrogen are required to react with 33.6 litres of oxygen to produce nitrogen dioxide.