10 L of 0.25 molof oxygen gas at a temperature of 30 °C was converted to ozone at the same temperature and pressure. Calculate the volume of ozone (O3) in mL? (Molar mass: O =16g/mol)

Question

3O2 → 2O3

Answer 1

First, we need to find the number of moles of oxygen gas we have:

n(O2) = (0.25 mol/L) x (10 L) = 2.5 mol

According to the balanced equation, 3 moles of oxygen gas produce 2 moles of ozone. Therefore, we can use a proportion to find the number of moles of ozone produced:

n(O3) = (2/3) x n(O2) = (2/3) x 2.5 mol = 1.67 mol

Now we can use the ideal gas law to find the volume of ozone:

PV = nRT

We know the pressure, temperature, and number of moles, and we can assume that the volume of ozone is essentially the same as the volume of oxygen gas that it was produced from. Therefore, we can rearrange the ideal gas law to solve for volume:

V = nRT/P

V = (1.67 mol) x (0.0821 L·atm/mol·K) x (303 K) / (1 atm)

V = 40.8 L

However, we want the answer in mL, so we need to convert:

V = 40.8 L x (1000 mL / 1 L) = 40,800 mL

Therefore, the volume of ozone produced is 40,800 mL (or 40.8 L).