Question

Aluminum reacts with chlorine gas to form aluminum chloride.

2Al(s)+3Cl2(g)→2AlCl3(s)

What minimum volume of chlorine gas (at 298 KK and 217 mmHgmmHg) is required to completely react with 8.48 gg of aluminum?

Answer 1

2. Al (5) 4 312 (9) → 2 Alclz (s) 2 mol of Ad reacts with 3 mol of che Number of moles of aluminum = molar mass 8.489 27 g/mol 0.314 mol 0.314 mol of Ad reacts with 3x0.314 = 0.471 mol of Cla pv=nRT I atm = 760 mm Hg) → Va HRT Р → Vs 0.471 molx 0.082 L-atmmolk-lx 298k 2 It 217 atm 760 0.471*0,082x298% 760 7 v= L 217 = v= 40.31 L & Minimum volume of Chlorine gas = 40.312