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Anonymous asked in Science & MathematicsChemistry · 8 months ago

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Predict the limiting reactant for each set of initial masses. How many grams of Cu(OH)2 are formed? CuCl2+ 2 NaOH ? Cu(OH)2 + 2 NaCl 5. a. b. C. 10g CuCl2 and 10g NaOH 20g CuCl2 and 5g NaOH 34.6g CuCl2 and 13.6g NaOH 6. Calculate the molarity (M) of each solution. a. b. c. 0.3mol of sodium hydroxide dissolved and diluted to 0.5L with water 14.6g of sodium hydroxide dissolved and diluted to 0.250L with water 24.6g of calcium chloride dissolved and diluted to 325mL with water

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  • 8 months ago

    5.

    Predict the limiting reactant for each set of initial masses. How many grams of Cu(OH)2 are formed?

    CuCl2+ 2 NaOH → Cu(OH)2 + 2 NaCl

    a.

    (10 g CuCl2) / (134.452 g CuCl2/mol) = 0.07438 mol CuCl2

    (10 g NaOH) / (39.99715 g NaOH/mol) = 0.2500 mol NaOH

    0.07438 mole of CuCl2 would react completely with 0.07438 x (2 mol NaOH / 1 mol CuCl2) = 0.14876 mole of NaOH, but there is more NaOH present than that, so NaOH is in excess and CuCl2 is the limiting reactant.

    b.

    (20 g CuCl2) / (134.452 g CuCl2/mol) = 0.1488 mol CuCl2

    (5 g NaOH) / (39.99715 g NaOH/mol) = 0.1250 mol NaOH

    0.1488 mole of CuCl2 would react completely with 0.1488 x (2 / 1) = 0.2976 mole of NaOH, but there is not that much NaOH present, so NaOH is the limiting reactant

    C.

    (34.6 g CuCl2) / (134.452 g CuCl2/mol) = 0.2573 mol CuCl2

    (13.6 g NaOH) / (39.99715 g NaOH/mol) = 0.3400 mol NaOH

    0.2573 mole of CuCl2 would react completely with 0.2573 x (2 / 1) = 0.5146 mole of NaOH, but there is not that much NaOH present, so NaOH is the limiting reactant

    6.

    Calculate the molarity (M) of each solution.

    a.

    (0.3 mol) / (0.5 L) = 0.6 mol/L = 0.6 M

    b.

    (14.6 g NaOH) / (39.99715 g NaOH/mol) / (0.250 L) = 1.46 mol/L = 1.46 M NaOH

    c.

    (24.6 g CaCl2) / (110.984 g CaCl2/mol) / (0.325 L) = 0.682 mol/L = 0.682 M CaCl2

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