1. a 5.455-g sample of impure cacl2 is dissolved and treated with excess potassium carbonate solution. the dried caco3 (calcium carbonate) precipitate weighs 4.010-g. calculate the percent by mass of cacl2 in the original mixture.

The mass percent of calcium chloride in the sample is 81.6 %

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of calcium carbonate = 4.010 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of calcium chloride and potassium carbonate follows:

By Stoichiometry of the reaction:

1 mole of calcium carbonate is produced from 1 mole of calcium chloride

So, 0.0401 moles of calcium carbonate will be produced from = of calcium chloride

Now, calculating the mass of calcium chloride by using equation 1:

Molar mass of calcium chloride = 111 g/mol

Moles of calcium chloride = 0.0401 moles

Putting values in equation 1, we get:

To calculate the mass percentage of calcium chloride in sample, we use the equation:

Mass of sample = 5.455 g

Mass of calcium chloride = 4.451 g

Putting values in above equation, we get:

Hence, the mass percent of calcium chloride in the sample is 81.6 %

i believe it is a.) released into

explanation:

the product of the pressure and volume is also constant