Question

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.

Answer 1

Answer:81.6%

Explanation:

Mass of CaCO3=4.010 g

Molar mass of CaCO3= 40+12+(16×3) = 100 g/mol.

Recall: number of moles(n)= mass÷ molar mass.

n=4.010÷100 = 0.0401 mol.

Molar mass of CaCl2 = 40+71= 111 g/mol.

Number of mol of CaCl2 = 5.455÷111= 0.04914 g/mol.

Mass of CaCl2 = 0.0401 × 111 = 4.4511 g of CaCl2.

Percent by mass of CaCl2 = (4.4511÷5.455) × 100

= 0.815967 ×100 = 81.5967%

Approximately; 81.6%.

Answer 2

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

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of calcium carbonate = 4.010 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{4.010g}{100g/mol}=0.0401mol$

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

$CaCl_2+K_2CO_3\rightarrow CaCO_3+2KCl$

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 = $\frac{1}{1}\times 0.0401=0.0401mol$ 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:

$0.0401mol=\frac{\text{Mass of calcium chloride}}{111g/mol}\\\\\text{Mass of calcium chloride}=(0.0401mol\times 111g/mol)=4.451g$

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

$\text{Mass percent of calcium chloride}=\frac{\text{Mass of calcium chloride}}{\text{Mass of sample}}\times 100$

Mass of sample = 5.455 g

Mass of calcium chloride = 4.451 g

Putting values in above equation, we get:

$\text{Mass percent of calcium chloride}=\frac{4.451g}{5.455g}\times 100=81.6\%$

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