Questions: Suppose a chemical reaction generated a 50% nitrogen / 50% oxygen (by volume) mixture of gas that had a total volume of 22.4 liters at STP. This gas sample is composed of: - 28 mole of nitrogen and 32 mole of oxygen. - 1 mole of nitrogen and 1 mole of oxygen. - 50 mole of nitrogen and 50 mole of oxygen. - 0.5 mole of nitrogen and 0.5 mole of oxygen. - none of the above

Solution Steps

We are given a gas mixture of nitrogen and oxygen with a total volume of 22.4 liters at standard temperature and pressure (STP). The mixture is 50% nitrogen and 50% oxygen by volume. We need to determine the number of moles of each gas in the mixture.

At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can use this information to find the number of moles of the gas mixture.

Since the total volume of the gas mixture is 22.4 liters, and 1 mole of gas occupies 22.4 liters at STP, the total number of moles of the gas mixture is: \[ \text{Total moles} = \frac{22.4 \text{ liters}}{22.4 \text{ liters/mole}} = 1 \text{ mole} \]

Given that the mixture is 50% nitrogen and 50% oxygen by volume, the moles of nitrogen and oxygen are each half of the total moles: \[ \text{Moles of nitrogen} = \frac{1 \text{ mole}}{2} = 0.5 \text{ mole} \] \[ \text{Moles of oxygen} = \frac{1 \text{ mole}}{2} = 0.5 \text{ mole} \]

Final Answer