Consider the following chemical reaction: 2SO2 (g) + O2 (g) > 2SO3 (g)
1.50 L. of sulfur trioxide at the pressure of 1.20 atm. and temperature of 25 oC is mixed with excess of oxygen.
Calclate volume of the product in L. at STP.
A. 11.2 L.
B. 1.65 L.
C. 16.5 L.
D. 0.129 L.

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

lol for the pointsss

it is the most commonly accepted one