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.