What volume of carbon dioxide gas (at STP) is produced when 1.01 moles of oxygen gas react completely?
2CO + O2 -> 2CO2
1 mol O2 gas @ STP gives 2 mols CO2 gas. One mole occupies 22.4 L @ STP so 2 mols will occupy ......?
To determine the volume of carbon dioxide gas produced when 1.01 moles of oxygen gas react completely, you can use the balanced chemical equation and the ideal gas law.
Step 1: Write the balanced chemical equation:
2CO + O2 -> 2CO2
Step 2: Identify the stoichiometric ratio:
From the balanced chemical equation, you can see that 1 mole of O2 reacts to produce 2 moles of CO2.
Step 3: Convert moles of O2 to moles of CO2:
Since we have 1.01 moles of O2, we can multiply it by the stoichiometric ratio (2 moles of CO2 / 1 mole of O2) to get the moles of CO2 produced:
1.01 moles O2 × (2 moles CO2 / 1 mole O2) = 2.02 moles CO2
Step 4: Use the ideal gas law equation to calculate the volume of CO2:
The ideal gas law equation states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At standard temperature and pressure (STP), the conditions are defined as 1 atm (101.325 kPa) and 273.15 K (0 °C).
From the ideal gas law equation, we can rearrange it to solve for volume:
V = (nRT) / P
Substituting the given values:
V = (2.02 moles × 0.0821 L·atm/(K·mol) × 273.15 K) / 1 atm
By performing the calculation, we find that the volume of carbon dioxide gas produced at STP is approximately 45.3 liters (rounded to the nearest tenth).