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

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it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.