What volume of oxygen and nitric oxide (2) (l) are required to obtain 118 l of nitric oxide (4)?

1. Let’s write down the reaction equation:

2NO + O2 = 2NO2.

2. Let’s convert the volume of nitrogen dioxide into the amount (Vm – molar volume, constant equal to 22.4 mol / l):

n (NO2) = V (NO2) / Vm = 118 L / 22.4 mol / L = 5.27 mol.

3. Using the equation, we find the amount of starting substances:

n (NO) = n (NO2) = 5.27 mol.

n (O2) = 0.5n (NO2) = 2.63 mol.

4. Let’s find the volumes of these substances:

V (NO) = n (NO) * Vm = 5.27 mol * 22.4 mol / L = 118 L.

V (O2) = n (O2) * Vm = 2.63 mol * 22.4 mol / l = 59 liters.

Answer: V (NO) = 118 l; V (O2) = 59 HP