chemistry question

Question

A sample of a mixture containing CaCl₂ and MgCl₂ weighed 2.000 g.The sample was dissolved in water, and H₂SO₄ was added until the precipitation of CaSO₄ was complete.

CaCl₂(aq) + H₂SO₄(aq) → CaSO₄(s) + 2HCl(aq)

The CaSO₄ was filtered, dried completely, and weighed. A total of 0.310 g of CaSO₄ was obtained.

How many moles of Ca²⁺ were in the CaSO₄?

n(Ca²⁺) = 

How many moles of Ca²⁺ were in the original 2.000-g sample?

n(Ca²⁺) =  

How many moles of CaCl₂ were in the 2.000-g sample?

n(CaCl₂) =  

How many grams of CaCl₂ were in the 2.000-g sample?

m(CaCl₂) =  

What was the percentage by mass of CaCl₂ in the original mixture?

w(CaCl₂) = 

What was the percentage by mass of MgCl₂ in the original mixture?

w(MgCl₂) = 

Answer 1

CaCl₂(aq) + H₂SO₄(aq) → CaSO₄(s) + 2HCl(aq)

How many moles of Ca²⁺ were in the CaSO₄?

0.310 g CaSO₄ x 1 mol/136.14 g = 2.28x10^-3 moles

How many moles of Ca²⁺ were in the original 2.000-g sample?

2.28x10^-3 moles since all of the calcium that is precipitated came from the CaCl₂ in the original sample

How many moles of CaCl₂ were in the 2.000-g sample?

2.28x10^-3 moles of CaCl₂ since each mole of Ca²⁺ comes from 1 mole of CaCl₂

How many grams of CaCl₂ were in the 2.000-g sample?

110.98 g CaCl2/mole x 2.28x10^-3 moles = 0.253 g

What was the percentage by mass of CaCl₂ in the original mixture?

0.253 g/2.000 g (x100%) = 12.7%

2.000 g - 0.253 g = 1.747 g MgCl2/2.000 g (x100%) = 87.4%