What weight of barium chloride will react with 2.36g of sodium sulphate in solution so as to produce 3.88g of barium sulphate and 1.94g of s

Question

What weight of barium chloride will react with 2.36g of sodium sulphate in solution so as to produce 3.88g of barium sulphate and 1.94g of sodium chloride in solution?

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Mộc Miên 4 years 2021-07-26T21:51:28+00:00 1 Answers 302 views 0

Answers ( )

  1. thienhuong
    0
    2021-07-26T21:53:20+00:00
    Reply

    Answer:

    3.46 g of BaCl2

    Explanation:

    The equation of the reaction is;

    BaCl2(aq) + Na2SO4(aq) ——–> 2NaCl(aq) + BaSO4(s)

    Number of moles of barium sulphate produced = 3.88 g/233.38 g/mol = 0.0166 moles

    From the reaction equation;

    1 mole of BaCl2 yields 1 mole of barium sulphate

    Hence 0.0166 moles of BaCl2 yields 0.0166 moles of barium sulphate

    Hence;

    Mass of BaCl2 required = 0.0166 moles × 208.23 g/mol = 3.46 g of BaCl2

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