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what minimum volume of chlorine gas (at 298 K and 237 mmHg) is required to completely...

what minimum volume of chlorine gas (at 298 K and 237 mmHg) is required to completely react with 7.44 g of aluminum?
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Answer #1

     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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