Answer :
To find the volume that 0.680 mol of oxygen gas occupies at standard temperature and pressure (STP), we can use the ideal gas law. However, at STP, we have a standard value for the volume occupied by one mole of any gas. This value is 22.4 liters per mole. Given this, we can calculate the volume of oxygen gas at STP by following these steps:
Step 1: Identify the number of moles of oxygen gas. We have 0.680 moles.
Step 2: Identify the volume that one mole of gas occupies at STP. It is 22.4 liters.
Step 3: Multiply the number of moles by the volume per mole to get the total volume.
[tex]\[ \text{Volume of gas at STP} = \text{Number of moles} \times \text{Volume per mole at STP} \][/tex]
[tex]\[ \text{Volume of oxygen gas} = 0.680 \text{ mol} \times 22.4 \text{ L/mol} \][/tex]
[tex]\[ \text{Volume of oxygen gas} = 15.232 \text{ L} \][/tex]
Therefore, 0.680 mol of oxygen gas occupies 15.232 liters at STP.
Step 1: Identify the number of moles of oxygen gas. We have 0.680 moles.
Step 2: Identify the volume that one mole of gas occupies at STP. It is 22.4 liters.
Step 3: Multiply the number of moles by the volume per mole to get the total volume.
[tex]\[ \text{Volume of gas at STP} = \text{Number of moles} \times \text{Volume per mole at STP} \][/tex]
[tex]\[ \text{Volume of oxygen gas} = 0.680 \text{ mol} \times 22.4 \text{ L/mol} \][/tex]
[tex]\[ \text{Volume of oxygen gas} = 15.232 \text{ L} \][/tex]
Therefore, 0.680 mol of oxygen gas occupies 15.232 liters at STP.
