How many grams of barium sulfate can be produced from .100 mole of barium Chloride?

in this reaction: BaCl2 + Na2SO4 ---> BaSO4 + 2NaCl

Question: How many grams of barium sulfate can be produced from 0.100 mole of barium chloride?

Can you please explain the process for me? I know there are 233.3906 g/mole of BaSO4 and .100 mole of BaCl2 is: 2028.324g. Please help.

2 Answers

  • by the equation:

    BaCl2 + Na2SO4 ---> BaSO4 + 2NaCl

    0.100 mole of BaCl2 makes an equal number of moles of BaSO4 = 0.100 moles of BaSO4

    using molar mass:

    0.100 moles of BaSO4 @ 233.39 g/mole of BaSO4 = 23.339 g BaSO4

    your answer rounded to 3 sig figs is 23.3 g BaSO4

  • 3BaCl2 + Fe2(SO4)3 ---> 2FeCl3 + 3Ba(SO4) So 0.a million moles of Barium Chloride reacts in basic terms with 0.033 moles of ferric sulfate (study restricting reagent concept given in references) to furnish 0.066 moles of Ferric Chloride and 0.a million moles of Barium Sulfate. as a consequence the mass of 0.a million moles of BaSO4 = 0.a million X 233.40 3 = 23.343 g.

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