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27 October, 10:55

A 8.85 g sample of hydrate of sodium carbonate, Na2CO3 ∙ XH2O was heated carefully until no more mass was lost from the sample. After heating, the final weight of the material was 7.55 g. How many moles of water left the compound? 0 moles

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  1. 27 October, 12:42
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    The compound is Na2CO3*1H2O

    There left 0.0721 moles H2O

    Explanation:

    Step 1: Data given

    Mass of Na2CO3 * XH2O = 8.85 grams

    After heating the mass = 7.55 grams

    Molar mass of H2O = 18.02 g/mol

    Molar mass of Na2CO3 = 105.99 g/mol

    Step 2: Calculate mass of water lost

    8.85 - 7.55 = 1.30 grams

    Step 3: Calculate moles Na2CO3

    Moles Na2CO3 = mass Na2CO3 / molar mass Na2CO3

    Moles Na2CO3 = 7.55 grams / 105.99 g/mol

    Moles Na2CO3 = 0.0712 moles

    Step 4: Calculate moles H2O

    Moles H2O = 1.30 grams / 18.02 g/mol

    Moles H2O = 0.0721

    Step 5: Calculate mol ratio

    We divide by the smallest amount of moles

    Na2CO3: 0.0712 / 0.0712 = 1

    H2O : 0.0721 / 0.0712 = 1

    This means for 1 mol Na2CO3 we have 1 mol H2O

    The compound is Na2CO3*1H2O

    There left 0.0721 moles H2O
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