Question

A 0.4584 g sample of impure zinc chloride, ZnCl2, FW = 136.3 g/mole, was dissolved in water and excess silver nitrate was added to the solution. The mass of the dry silver chloride, AgCl, FW = 143.3 g/mole, produced was 0.8564 g. The balanced precipitation is: ZnCl2 (aq) + 2 AgNO3 (aq) --> 2 Agcl (s) + Zn(NO3)2 (aq).

1. Calculate the moles of zinc chloride present in the sample.

2. Calculate the mass of zinc chloride in the sample.

3. Calculate the percent purity of the zinc cholride in the sample.

Answer 1

A 0.4584 g sample of impure zinc chloride, ZnCl2, FW = 136.3 g/mole, was dissolved in water and excess silver nitrate was added to the solution. The mass of the dry silver chloride, AgCl, FW = 143.3 g/mole, produced was 0.8564 g. The balanced precipitation is: ZnCl2 (aq) + 2 AgNO3 (aq) --> 2 Agcl (s) + Zn(NO3)2 (aq).

1. Calculate the moles of zinc chloride present in the sample.
<b>Convert mass AgCl to moles AgCl. Using the coefficients in the balanced equation, convert moles AgCl to moles ZnCl2.</b>

2. Calculate the mass of zinc chloride in the sample.
<b>g ZnCl2 = moles ZnCl2 x molar mas ZnCl2.</b>

3. Calculate the percent purity of the zinc chloride in the sample.
<b>%ZnCl2 = (mass ZnCl2/mass sample)*100 = ?</b>