Question

A mixture consisting of only zinc(II) chloride (ZnCl2, 136.31 g/mol) and titanium(IV) chloride (TiCl4, 189.67 g/mol) weighs 1.0971 g. When the mixture is dissolved in water and an excess of silver acetate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl, 143.32 g/mol). The mass of the silver chloride is found to be 2.9691 g. Calculate the mass percentage of zinc(II) chloride in the original mixture.

Answer 1

Suppose the mixture contains x grams of ZnCl₂ and y grams of TiCl₄ .

SO, x + y = 1.0971 .....................................................................Equation 1

Moles of ZnCl₂ present : mass / molar mass of ZnCl₂ = x g/ (136.31 g/mol) = 0.00734x mole

The reaction between ZnCl₂ and silver acetate is given below:

ZnCl₂ (aq) + 2 AgOCOCH₃ (aq) = Zn(OCOCH₃)₂ (aq) + 2 AgCl (s)

From the balanced stoichiometry, we get:

Moles of AgCl : Moles of ZnCl₂ = 2:1

So, moles of AgCl = 2 * moles of ZnCl₂

So, in the reaction moles of AgCl produced = 2* 0.00734x moles = 0.01468x mole

Moles of TiCl₄ present : mass / molar mass of TiCl₄ = y g/ (189.67 g/mol) = 0.00527y mole

Similarly the reaction between TiCl₄ and silver acetate is:

TiCl₄ (aq) + 4 AgOCOCH₃ (aq) = Ti(OCOCH₃)₄ (aq) + 4 AgCl (s)

From the balanced stoichiometry, we get:

Moles of AgCl : Moles of TiCl₄ = 4:1

So, moles of AgCl = 4 * moles of TiCl₄

So, in the reaction moles of AgCl produced = 4* 0.00527y moles = 0.02108y mole

Total moles of AgCl produced = 0.01468x mole + 0.02108y mole

Mass of AgCl produced = Moles * molar mass of AgCl = (0.01468x + 0.02108y) * 143.32 g

According to the question, mass of AgCl produced = 2.9691 g

So, (0.01468x + 0.02108y) * 143.32 g = 2.9691

Or, 0.01468x + 0.02108y = 0.0207

Substituting, y = 1.0971 -x from Equation 1 in the above equation, we get:

0.01468x + 0.02108* ( 1.0971 -x ) = 0.0207

Or, - 0.0064 x = - 0.0024

So, x = 0.375

So, in the mixture mass of ZnCl2 is 0.375 g

The mass percentage of ZnCl2 is : (0.375/ 1.0971) * 100% = 34.18%