Question

What volume of sulfur dioxide gas at 45°C and 723 mmHg will react

completely with 0.870 L of oxygen gas at constant temperature and
pressure? R = 0.08206 Lxatm/Kxmol
2SO₂(g) + O₂(g) → 2503(g)
A. 1.74L
B. 0.0317L
C. 0.870L
D. 0.00634L
E. 3.48L"

Answer 1

The volume of sulfur dioxide gas at 45°C and 723 mmHg that will react completely with 0.870 L of oxygen gas is 0.0317 L.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:

• P is the pressure of the gas
• V is the volume of the gas
• n is the number of moles of the gas
• R is the ideal gas constant
• T is the temperature of the gas

We know the volume of oxygen gas (0.870 L) and the temperature and pressure of sulfur dioxide gas (45°C and 723 mmHg). To find the volume of sulfur dioxide gas, we can rearrange the ideal gas law equation:

V1/V2 = P1/P2

Plugging in the values:

V2 = (0.870 L) x (723 mmHg/723 mmHg) x (45°C/273 K) = 0.0317 L

Therefore, the volume of sulfur dioxide gas is 0.0317 L.