How do I solve the following? Using your answer for Question 2, determine the ma

Question

How do I solve the following?

Using your answer for Question 2, determine the mass of potassium carbonate needed to fully precipitate all the calcium from a 25 mL sample of 15% calcium chloride solution. (Note: this is the same concentration as the solution you used in Experiment

My answer to question 2 was:

15 g CaCl2 (1 mole CaCl2/110.98 g CaCl2) = 0.135 moles CaCl2

15 g K2CO3 (1mole K2CO3/138.21 g K2CO3) = 0.1085 moles K2CO3

                So limiting reagent is K2CO3

I'm completely lost...thanks in advance :)

Explanation / Answer

given 25 ml of 15% Calcium chloride solution .By using wieght volume percentage

Another variation on percentage concentration is weight/volume percent or mass/volume percent. This variation measures the amount of solute in grams but measures the amount of solution in milliliters. An example would be a 5%(w/v) NaCl solution. It contains 5 g of NaCl for every 100. mL of solution.

weight of solute (in g)
volume of solution (in mL)

100 ml of the solution caontains 15 g of calcium chloride

then 25 ml of solution contains (25*15/100)= 3.75 g of Calcium Chloride.

Moles of CaCl2 = 3.75/110.98=0.0338 moles

the balanced reaction is

K2CO3 + CaCl2 = CaCO3 + 2KCl

So 1 mole of CaCl2 requires 1 mole of K2CO3 for complete precpitation

therefore 0.0338 moles of CaCl2 requires 0.0338 moles of K2CO3

mass of K2CO3 required= 0.0338*138.21=4.67 g

weight of solute (in g)
volume of solution (in mL)

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