Question 10.4.4: Use gas laws in stoichiometry calculations. A sample of O2 w......
Use gas laws in stoichiometry calculations.
A sample of O_{2} with a pressure of 1.42 atm and a volume of 250. mL is allowed to react with excess SO_{2} at 129 ºC.
2 SO_{2}(g) + O_{2}(g) → 2 SO_{3}(g)
Calculate the pressure of the SO_{3} produced in the reaction if it is transferred to a 1.00-L flask and cooled to 35.0 ºC.
You are asked to calculate the pressure of a gas produced in a chemical reaction.
You are given the balanced equation for the reaction; the pressure, volume, and temperature of a reactant; and the volume and temperature of the gas produced in the reaction.
Step 1. Calculate the amount of reactant (O_{2}) available using the ideal gas law.
n_{o_{2}} = \frac{PV}{RT} = \frac{\left(1.42 \text{ atm}\right) \left(0.250 \text{ L}\right) }{\left(0.082057 \text{L} \cdot \text{atm/K} \cdot \text{mol}\right) \left(129 + 273.15\text{ K}\right) } = 0.0108 mol
Step 2. Use the amount of limiting reactant (O_{2}) and the balanced equation to calculate the amount of SO_{3} produced.
0.0108 oml O_{2} × \frac{2\text{ mol SO}_{3}}{1\text{ mol O}_{2}} = 0.0216 mol SO_{3}
Step 3. Use the ideal gas law and the new volume and temperature conditions to calculate
the pressure of SO_{3}.
P_{SO_{3}} = \frac{nRT}{V} = \frac{\left(0.0216 \text{ mol SO}_{3}\right)\left(0.082057 \text{ L} \cdot \text{atm/K} \cdot \text{mol} \right) (35.0 + 273.15 \text{ K})}{1.00\text{ L}} = 0.546 atm