Question #ba0cd

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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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