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23 December, 15:16

If an experiment with 10.2 g barium chloride produced 14.5 g silver chloride, calculate thr experimental mole ratio of silver chloride to barium chloride

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  1. 23 December, 15:31
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    The experimental mole ratio of silver chloride to barium chloride is calculated as below

    fin the mole of each compound

    mole = mass/molar mass

    moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl

    moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2

    find the mole ratio by dividing each mole with the smallest mole (0.049)

    AgCl = 0.102/0.049 = 2

    BaCl2 = 0.049/0.049 = 1

    therefore the mole ratio AgCl to BaCl2 is 2 : 1
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