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13 March, 11:12

What is the amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol and Al2S3 (s) + 6 H2O (l) → 2 Al (OH) 3 (s) + 3 H2S (g)

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  1. 13 March, 11:28
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    From the equation, we can tell that 1 mol of Al₂S₃ requires 6 moles of water.

    The molar ratio is 1/6

    Moles of Al₂S₃ present = 20/150.17

    = 0.133

    Moles of water present = 2/18.02

    = 0.111

    The moles of Al₂S₃ that will react are:

    0.111/6

    = 0.0185

    The remaining amount:

    0.133 - 0.0185

    = 0.1145 mol

    Or

    0.1145 * 150.17

    = 17.19 grams
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