A 10.00 g sample of mixture containing NaCl and K_2SO_4 is dissolved in water. T

Question

A 10.00 g sample of mixture containing NaCl and K_2SO_4 is dissolved in water. This aqueous mixture then reacts with excess Pb(NO_3)_2 (aq) to form a 21.75 g mixture of solid containing PbCl_2(s) and PbSO_4(s). Determine the mass percentage NaCl in the original mixture. Which applies here, "ion-partner exchange" or combustion? Which of the following applies: a single reaction, or parallel reactions? If parallel reactions are involved, can they be added together in doing stoichiometric calculations? What can you define as unknown variables? How will they be used to determine the mass Percent of NaCl? What equations represent (a) the total mass of NaCl and K_2SO_4 in the reactant mixture, and (b) the total mass of PbCl_2(s) and PbSO_4(s) in the product mixture, in terms of your defined unknown?

Explanation / Answer

The lead(II) ions for solids with both the chloride ions and sulfate ions .

). The reactions are:

Pb2+ (aq) + 2Cl- (aq) PbCl2 (s)

Pb2+ (aq) + SO4 2- (aq) PbSO4 (s)

We want to determine the percent by mass sodium chloride, which is

( mass NaCl/ mass mixture) x 100% = ( mass NaCl /10.00 g) x 100%

So, we need to know the mass of NaCl.

Let x = mass of NaCl and let 10.00-x = mass K2SO4.

molar mass of NaCl = 58.44 g/mol

molar mass of K2SO4 = 174.27 g/mol

molar mass of PbCl2 = 278.1 g/mol

molar mass of PbSO4= 303.27 g/mol

x /58.44 = moles NaCl

and 10 – x/174.27 = moles K2SO4

mass of PbCl2 + mass PbSO4 = total mass of solid (278.1)

(278.1)(½) [x/ 58.44] x + (303.27) [10 - x /174.27] = 21.75

x = 6.807 g mass NaCl

(mass NaCl /10.00 g) * 100% =6.807 g/10 g) x 100% = 68.07% NaCl by mass--------------answer

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