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valentinak56 [21]
3 years ago
13

Calcium chloride (aq) reacts with sodium carbonate (aq) to from solid calcium carbonate and aqueous sodium chloride. Determine t

he 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!)
Chemistry
2 answers:
sattari [20]3 years ago
8 0

Answer:

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

natulia [17]3 years ago
4 0

Answer:

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

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How many significant figures does 2.0 x 10² have?​
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Suppose now that you wanted to determine the density of a small crystal to confirm that it is sulfur. From the literature, you k
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Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so <u>20 mL of the mixture would weigh</u>:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

   1. Express Y in terms of X:

X + Y = 20

  • Y = 20 - X

   2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

  • 1.492*X + 2.890*(20-X) = 41.4

   3. Solve for X:

  • 1.492*X + 57.8 - 2.890*X = 41.4
  • 1.398*X = 16.4
  • X = 11.73 mL

   4. Using the now known value of X, solve for Y:

X + Y = 20

  • 11.73 + Y = 20
  • Y = 8.27 mL

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of  CHCl₃.

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