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Calculate the volume in ) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO 3 (s) . There is an excess of CaCl 2. What’s the volume of sodium carbonate?

Calculate the volume in of 0100 M Na2CO3 needed to produce 100 g of CaCO 3 s There is an excess of CaCl 2 Whats the volume of sodium carbonate class=

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Eduard22sly

Answer:

100 mL of Na2CO3

Explanation:

We'll begin by calculating the number of mole in 1 g of CaCO3. This can be obtained as follow:

Mass of CaCO3 = 1 g

Molar mass of CaCO3 = 100.09 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 1/100.09

Mole of CaCO3 = 0.01 mole

Next, we shall determine the number of mole of Na2CO3 needed to produce 0.01 mole of CaCO3.

This is illustrated below:

Na2CO3 + CaCl2 —> 2NaCl + CaCO3

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CaCO3.

Therefore, 0.01 mole of Na2CO3 will also react to produce 0.01 mole of CaCO3.

Next, we shall determine the volume of Na2CO3 needed for the reaction as illustrated below:

Mole of Na2CO3 = 0.01 mole

Molarity of Na2CO3 = 0.1 M

Volume of Na2CO3 solution needed =?

Molarity = mole /Volume

0.1 = 0.01 / volume of Na2CO3

Cross multiply

0.1 × volume of Na2CO3 = 0.01

Divide both side by 0.1

Volume of Na2CO3 = 0.01 / 0.1

Volume of Na2CO3 = 0.1 L

Finally, we shall convert 0.1 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.1 L = 0.1 L × 1000 mL / 1 L

0.1 L = 100 mL

Thus, 0.1 L is equivalent to 100 mL.

Therefore, 100 mL of Na2CO3 is needed for the reaction.

The volume of sodium carbonate used in the given reaction is 0.1 liter or 100mL.

How we calculate the volume from molarity?

Volume of the solution will be calculated by using the molarity as:

M = n/V, where

n is the moles of solute and this can be calculated as:

n = W/M , where

W = given or required mass

M = molar mass

Given chemical reaction is:
Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl

Moles of 1g of CaCO₃ = 1g/100g/mole = 0.01 moles

From the stoichiometry of the reaction, it is clear that same moles of CaCO₃ produced by same moles of used Na₂CO₃.

0.01 moles of CaCO₃ = produced by 0.01 moles of Na₂CO₃

Now we calculate the volume of 0.01 moles and 0.100M of Na₂CO₃ as:

V = n/M

V= 0.01/0.100 = 0.1L or 100mL

Hence 0.1 L is the required volume.

To know more about molarity, visit the below link:
https://brainly.com/question/16343005

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