Question

How many grams of zinc phosphate are formed when 10.0 grams of zinc react with excess phosphoric acid

Answer 1

Task:

How many grams of zinc phosphate are formed when 10.0 grams of zinc react with excess phosphoric acid

Solution:

The chemical equation for this reaction is

$$3 \mathrm{Zn}(\mathrm{s}) + 2 \mathrm{H_3PO_4}(\mathrm{aq}) = \mathrm{Zn_3(PO_4)_2}(\mathrm{aq}) + 3 \mathrm{H_2}(\mathrm{g})$$

The molar weight of zinc phosphate is

$$\mathrm{MW}(\mathrm{Zn_3(PO_4)_2}) = 3 \cdot \mathrm{MW}(\mathrm{Zn}) + 2 \cdot \mathrm{MW}(\mathrm{P}) + 8 \cdot \mathrm{MW}(\mathrm{O})$$ $$\mathrm{MW}(\mathrm{Zn_3(PO_4)_2}) = 3 \cdot 65 + 2 \cdot 31 + 8 \cdot 16 = 385 \ \mathrm{g/mol}$$

The number of moles of Zn is

$$n(\mathrm{Zn}) = \frac{m(\mathrm{Zn})}{MW(\mathrm{Zn})} = \frac{10.0}{65} = 0.154 \ \mathrm{mol}$$

The number of moles of $\mathrm{Zn_3(PO_4)_2}$ is

$$n(\mathrm{Zn_3(PO_4)_2}) = \frac{n(\mathrm{Zn})}{3} = \frac{0.154}{3} = 0.051 \ \mathrm{mol}$$

The mass of $\mathrm{Zn_3(PO_4)_2}$ is

$$m(\mathrm{Zn_3(PO_4)_2}) = \frac{n(\mathrm{Zn_3(PO_4)_2})}{MW(\mathrm{Zn_3(PO_4)_2})} = \frac{0.051}{385} = 19.6 \ \mathrm{g}$$

Answer: $m(\mathrm{Zn_3(PO_4)_2}) = 19.6 \ \mathrm{g}$