Question

A 110.0g sample of a mixture of CaCl2 and NaCl is treated with Na2CO3 to precipitate the Calcium as Calcium Carbonate. This CaCO3 is heated to convert all the Calcium into CaO and the final mass of CaO is 11.62 grams. The % by mass of the CaCl2 in the original mixture has to be found out..

I've tried out finding the no of moles of CaCO3 which which i got as 0.2075 and assuming that the CaCO3 and the no of moles of CaCl2 should be equal i tried finding out the % which came out to be 20.75% in this method and if i separately found the masses of Ca, Cl2 and summed them up to find the percentage over 110 the % turned out to be 20.09% need some suggestions to get the % anywhere near 15.2% in this process taking all of the compounds to be anhydrous

Answer 1

To find the percentage by mass of CaCl₂ in the original mixture, you need to use stoichiometry and the mass of CaO obtained after heating the CaCO3. The percentage of CaCl2 in the original mixture is approximately 19.98%.

Step-by-step explanation:

To find the percentage by mass of CaCl₂in the original mixture, we need to use stoichiometry and the mass of CaO obtained after heating the CaCO3.

First, calculate the number of moles of CaO obtained:

Number of moles of CaO = Mass of CaO / molar mass of CaO = 11.62 g / 56.08 g/mol = 0.2073 mol

Since the reaction between CaCO3 and Na2CO3 occurs in a 1:1 ratio, the number of moles of CaCl₂ in the original mixture is also 0.2073 mol. So, to find the percentage by mass of CaCl₂:

Percentage of CaCl₂ = (mass of CaCl₂ / mass of original mixture) x 100 = (0.2073 mol x 110.98 g/mol) / 110.0 g x 100 = 19.98%

Therefore, the percentage by mass of CaCl₂ in the original mixture is approximately 19.98%.