To find the volume of oxygen gas produced in the synthesis of ethanol, we need to first calculate the number of moles of ethanol produced from the given mass of ethanol. Next, we use the mole ratio from the balanced equation to find the number of moles of oxygen gas. Finally, we use the ideal gas law to calculate the volume of oxygen gas. Volume of O₂ gas ≈ 7.8 L (rounded to 2 decimal places).
The balanced equation for the synthesis of ethanol is:
2 CO (g) + 3 H₂ (g) → CH₃CH₂OH (g) + 1/2 O₂ (g)
To find the volume of oxygen gas produced, we need to use the stoichiometric coefficients from the balanced equation. The coefficient in front of O₂ is 1/2, which means that for every 1 mole of ethanol produced, 1/2 mole of O₂ is required.
Since we know the mass of ethanol produced (23 g), we can use its molar mass (46.07 g/mol) to calculate the number of moles. Then, we can use the mole ratio to find the number of moles of O₂. Finally, we can use the ideal gas law to calculate the volume of O₂ gas.
First, calculate the number of moles of ethanol:
moles of ethanol = mass of ethanol / molar mass of ethanol
= 23 g / 46.07 g/mol
= 0.4996 mol (rounded to 4 decimal places)
Next, use the mole ratio to find the number of moles of O₂:
moles of O₂ = moles of ethanol × (1/2)
= 0.4996 mol × (1/2)
= 0.2498 mol (rounded to 4 decimal places)
Now, we can use the ideal gas law to find the volume of O₂ gas:
volume of O₂ gas = moles of O₂ × (R × temperature / pressure)
Convert the temperature to Kelvin:
temperature in Kelvin = 325 K
Plug in the values into the ideal gas law equation:
volume of O₂ gas = 0.2498 mol × (0.0821 atm·L/mol·K × 325 K / 1.03 atm)
Solve for the volume of O₂ gas:
volume of O₂ gas ≈ 7.8 L (rounded to 2 decimal places)