Aluminum reacts with chlorine gas to form aluminum chloride. 2al (s) + 3cl2 (g) →2alcl3 (s) what minimum volume of chlorine gas (at 298 k and 225 mmhg) is required to completely react with 7.85 g of aluminum

The balanced chemical reaction is expressed as:

2Al (s) + 3Cl2 (g) →2AlCl3 (s)

To determine the volume of chlorine gas needed given the mass of aluminum metal to be used, we need to calculate for the moles of chlorine needed and use a relation that relates moles and volume by assuming the gas to be an ideal gas. We use the equation PV = nRT. We calculate as follows:

7.85 g Al (1 mol / 26.98 g) (3 mol Cl2 / 2 mol Al) = 0.43643 mol Cl2

PV = nRT

V = nRT / P

V = 0.43643 (0.08205) (298) / (225/760)

V = 36.04 L chlorine gas

The minimum volume needed would be 36.04 L.