# Question ba0cd

Mar 1, 2017

$46 \text{g}$ of "Zn"_3("PO"_4)_2 can be produced (theoretically).

#### Explanation:

According to the chemical equation, 3 mol of ${\text{ZnCl}}_{2}$ will combine with 2 mol of ${\text{K"_3"PO}}_{4}$ to produce 1 mol of "Zn"_3("PO"_4)_2 (and 6 mol of $\text{KCl}$). (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

• ${\text{ZnCl}}_{2}$: 136.4 g/mol
• ${\text{K"_3"PO}}_{4}$: 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

$\left(n \text{ZnCl"_2) / (n"K"_3"PO"_4) = (0.7331 "mol")/(0.236 "mol}\right)$

$= 3.11 > 1.5$

This shows that ${\text{K"_3"PO}}_{4}$ is the limiting reagent while ${\text{ZnCl}}_{2}$ is in excess.

Since 2 mol of ${\text{K"_3"PO}}_{4}$ produces 1 mol of "Zn"_3("PO"_4)_2, 0.236 mol of ${\text{K"_3"PO}}_{4}$ produces

$0.236 \text{mol" xx 1/2 = 0.118 "mol}$

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

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

$0.118 \text{mol" xx 386.2 "g/mol" = 46 "g}$