2Al(s) + 3Cl2(g) ----> 2AlCl3(s)
no of mol of Al reacted = 7.44/27 = 0.275 mol
no of mol of Cl2 required = 0.275*3/2 = 0.4125 mol
volume of Cl2 required = nRT/P
= 0.4125*0.0821*298/(237/760)
= 32.36 L
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