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7 July, 02:30

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 28.0 g of aluminum?

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  1. 7 July, 05:20
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    Answer: 2Al (s) + 3Cl2 (g) - - - > 2AlCl3 (s) Always work in moles moles = mass of substane / molar mass 2Al (s) + 3Cl (g) - - > 2AlCl3 (s) moles Al = 23.0 g / 26.98 g/mol = 0.852 moles Al
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