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24 December, 02:50

Calcium chloride (aq) reacts with sodium carbonate (aq) to from solid calcium carbonate and aqueous sodium chloride. Determine the volume of a 2.00 M Calcium chloride solution would be needed to exactly react with 0.0650 L of 1.50 M Na2CO3. (Use BCA!)

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  1. 24 December, 04:33
    0
    0.0488 L

    Explanation:

    Step 1:

    The balanced equation for the reaction. This is given below:

    CaCl2 (aq) + Na2CO3 (aq) - > CaCO3 (s) + 2NaCl (aq)

    Step 2:

    Determination of the number of mole of Na2CO3 in 0.0650 L of 1.50 M Na2CO3 solution.

    Volume of solution = 0.0650 L

    Molarity of Na2CO3 = 1.50 M

    Mole of solute (Na2CO3) = ?

    Molarity = mole of solute / Volume of solution

    1.50 = mole of solute/0.0650

    Cross multiply to express in linear form.

    Mole of solute = 1.5 x 0.0650

    Mole of solute (Na2CO3) = 0.0975 mole

    Step 3:

    Determination of the number of CaCl2 that reacted.

    CaCl2 (aq) + Na2CO3 (aq) - > CaCO3 (s) + 2NaCl (aq)

    From the balanced equation,

    1 mole of CaCl2 reacted with 1 mole Na2CO3.

    Therefore, 0.0975 mole of CaCl2 will also react with 0.0975 mole of Na2CO3.

    Step 4:

    Determination of the volume of CaCl2 that reacted.

    Mole of solute (CaCl2) = 0.0975 mole

    Molarity of CaCl2 = 2.00 M

    Volume of solution = ?

    Molarity = mole of solute / Volume

    2 = 0.0975/volume

    Cross multiply to express in linear form

    2 x Volume = 0.0975

    Divide both side by 2

    Volume = 0.0975/2

    Volume = 0.0488 L

    Therefore, the volume of CaCl2 that is 0.0488 L
  2. 24 December, 04:53
    0
    We need a volume of 48.75 mL of CaCl2 to react

    Explanation:

    Step 1: Data given

    Molarity of calcium chloride = 2.00 M

    Volume of Na2CO3 = 0.0650 L

    Molarity of Na2CO3 = 1.50 M

    Step 2: The balanced equation

    CaCl2 (aq) + Na2CO3 (aq) → CaCO3 + 2NaCl

    Step 3: Calcumate moles Na2CO3

    Moles Na2CO3 = molarity Na2CO3 * volume

    Moles Na2CO3 = 1.50 M * 0.0650 L

    Moles Na2CO3 = 0.0975 moles

    Step 4: Calculate moles CaCl2 neede to react

    For 1 mol CaCl2 we need 1 mol Na2CO3 to produce 1 mol CaCO3 and 2 moles NaCl

    For 0.0975 moles Na2CO3 we need 0.0975 moles CaCl2

    Step 5: Calculate volume of CaCl2 solution

    Volume = moles CaCl2 / molarity CaCl2

    Volume = 0.0975 moles / 2.00 M

    Volume = 0.04875 L = 48.75 mL

    We need a volume of 48.75 mL of CaCl2 to react
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Home » Chemistry » Calcium chloride (aq) reacts with sodium carbonate (aq) to from solid calcium carbonate and aqueous sodium chloride. Determine the volume of a 2.00 M Calcium chloride solution would be needed to exactly react with 0.0650 L of 1.50 M Na2CO3.
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