6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3

Question

6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3

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Philomena 4 years 2021-07-28T20:07:11+00:00 1 Answers 8 views 0

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  1. RobertKer
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    2021-07-28T20:08:32+00:00
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    Answer:

    23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr

    Explanation:

    The balanced equation here is

    6NaBr + 1AlO3 = 3Na2O + 2AlBr3

    6 moles of NaBr are required to produce 2 moles of AlBr3

    Mass of one mole of NaBr = 102.894 g/mol

    Mass of one mole of AlBr3 = 266.69 g/mol

    Mass of 6 moles of NaBr = 6*102.894 g/mol

    Mass of two moles of AlBr3 = 2*266.69 g/mol

    6*102.894 g  NaBr produces 2*266.69 g of AlBr3

    23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr

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