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Bio
Mar 1, 2017

46 "g" of "Zn"_3("PO"_4)_2 can be produced (theoretically).

Explanation:

According to the chemical equation, 3 mol of "ZnCl"_2 will combine with 2 mol of "K"_3"PO"_4 to produce 1 mol of "Zn"_3("PO"_4)_2 (and 6 mol of "KCl"). (Using the ratio of the masses would be a mistake!)

From the periodic table of elements, the atomic mass of the relevant elements are

  • Zn: 65.4
  • Cl: 35.5
  • K: 39.1
  • P: 31.0
  • O: 16.0

From the data, the molar masses of the species of interest are calculated to be

  • "ZnCl"_2: 136.4 g/mol
  • "K"_3"PO"_4: 212.3 g/mol
  • "Zn"_3("PO"_4)_2: 386.2 g/mol

We can thus find the amount of each reactants in mol.

n_("ZnCl"_2) = frac{100 "g"}{136.4 "g/mol"} = 0.7331 "mol"

n_("K"_3"PO"_4) = frac{50 "g"}{212.3 "g/mol"} = 0.236 "mol"

Ideally, the ratio of reactants would be 3/2 = 1.5. However, we have got

(n"ZnCl"_2) / (n"K"_3"PO"_4) = (0.7331 "mol")/(0.236 "mol")

= 3.11 > 1.5

This shows that "K"_3"PO"_4 is the limiting reagent while "ZnCl"_2 is in excess.

Since 2 mol of "K"_3"PO"_4 produces 1 mol of "Zn"_3("PO"_4)_2, 0.236 mol of "K"_3"PO"_4 produces

0.236 "mol" xx 1/2 = 0.118 "mol"

of "Zn"_3("PO"_4)_2.

To get the yield of the product in grams, multiply by its molar mass

0.118 "mol" xx 386.2 "g/mol" = 46 "g"

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