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HELP WITH CHEMISTRY PLEASE!!! (college level)?

1) If 710. mL of 0.04868 M aqueous NaBr and 1.134 g of gaseous Cl2 are reacted stoichiometrically according to the equation, how many mol of gaseous Cl2 remained?

2 NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(l)

2) If 570 mL of 0.0228 M aqueous BaCl2 and 860 mL of 0.00534 M aqueous Al2(SO4)3 are reacted, what mass (g) of solid BaSO4 is produced?

Al2(SO4)3(aq) + 3 BaCl2(aq) → 2 AlCl3(aq) + 3 BaSO4(s)

2 Answers

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  • 1 decade ago
    Favorite Answer

    1)

    .710 L ( 0.04868 M) = .0346 mol NaBr

    1.134 g Cl2 (1 mol / 70.9 g) = .016 mol Cl2

    For 1 mol Cl2, 2 mol NaBr react.

    .016 * 2 = .032 mol NaBr react with .016 mol Cl2

    THERE IS NO Cl2 REMAINING, THERE IS NaBr REMAINING.

    .0345 - .032 = 0.0025 mol NaBr remain.

    2)

    .570 L (.0228 M) = .013 mol BaCl2

    .860 L (.00534 M) = .0046 mol Al2(SO4)3

    For 1 mole Al2(SO4)3, 3 mol BaCl2 are needed.

    .0046 * 3 = .014 mol Al2(SO4)3 needed

    BaCl2 is limiting.

    .013 mol BaCl2

    The ratio of BaCl2 and BaSO4 is 1:1

    So, .013 mol of BaSO4 are produced.

    .013 mol BaSO4 ( 233.4 g / 1 mol ) =

    3.034 grams of BaSO4

    Source(s): =] My Brain.
  • ?
    Lv 4
    5 years ago

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