Question

Aqueous Na₂SO₃ may be used to remove SO₂(g) due to the reaction:

Na2SO₃(aq) + SO₂(g) + H₂O(l) → 2NaHSO₃(aq).
When aqueous Na2SO3 was placed in a room containing 3.5 × 104 L of SO₂-contaminated air at 25°C, the pressure changed from 0.9955 atm to 0.9900 atm. How many moles of SO₂ were removed by the chemical reaction with Na₂SO₃(aq)?
a) 0.24 mol
b) 0.36 mol
c) 0.48 mol
d) 0.60 mol

Answer 1

d) 0.60 mol. Approximately 0.60 moles of SO2 were removed by reacting with Na2SO3, calculated using the change in pressure, the volume of the air, and the ideal gas law equation.

Step-by-step explanation:

The problem asks how many moles of SO2 were removed from the air by the reaction with Na2SO3(aq). To find this, we apply the ideal gas law equation PV = nRT, where 'P' is the pressure in atm, 'V' is the volume in liters, 'n' is the number of moles, 'R' is the gas constant (0.0821 L·atm/mol·K), and 'T' is the temperature in Kelvin. The change in pressure (0.9955 atm to 0.9900 atm) represents the amount of SO2 that reacted with Na2SO3.

First, convert the change in pressure to moles of SO2 by rearranging the ideal gas law to solve for 'n' (n=P*V/(R*T)). The volume is 3.5 × 104 L and the temperature is 298 K (25°C = 298 K).

Now, calculate the moles of SO2:

Δn = (Pfinal - Pinitial) * V / (R * T) = (0.9955 atm - 0.9900 atm) * 3.5 × 104 L / (0.0821 L·atm/mol·K * 298 K)

After performing the calculation, you find that approximately 0.60 moles of SO2 were removed by Na2SO3, which corresponds to choice d