Specifically chemistry stemlords A 10.00 g sample of an unknown water soluble barium salt is treated with an excess of Na2SO4 to precipitate 11.21 g BaSO4. Which barium salt is it?

A. BaCl2

B. Ba(O2CH)2

C. Ba(NO3)2

D. BaBr2

Help

  • Dyno [he/him]
    ·
    2 years ago

    Mass = moles * Mr
    Moles = mass/Mr
    BaSO4 Mr = 233.43
    Moles BaSO4 = 11.21g / 233.43 = 0.048 mol
    Reaction is 1:1, thus 0.048 mol of initial salt produces 0.048 mol of resultant salt

    Mr = Mass / moles
    10.00 g / 0.048 = 208. 33
    Since Barium Mr is 137.33, 208.33 - 137.33 = 71

    Cl2 = 71
    (O2CH)2 = 90
    (NO3)2 = 124
    Br2 = 160

    Therefore BaCl2