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!)

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

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