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6 October, 11:58

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student generated 3.7 g of barium carbonate from 2.0 g of their alkali metal carbonate. M2CO3 (aq) + BaCl2 (aq) - -> 2MCl (aq) + BaCO3 (s) How many moles of alkali metal carbonate were reacted? Question 3 options: 0.019 mol 0.038 mol 0.094 mol 2.3 mol

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  1. 6 October, 12:27
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    0.019 moles of M2CO3

    Explanation:

    M2CO3 (aq) + BaCl2 (aq) - -> 2MCl (aq) + BaCO3 (s)

    From the equation above;

    1 mol of M2CO3 reacts to produce 1 mol of BaCO3

    Mass of BaCO3 formed = 3.7g

    Molar mass of BaCO3 = 197.34g/mol

    Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol

    Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

    1 = 1

    x = 0.019

    x = 0.019 moles of M2CO3
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