Question

In the following reaction, how many grams of potassium phosphate, K3PO4, will be produced from 62.3 g of potassium hydroxide, KOH?

3KOH + H3PO4 ---> K3PO4 +3H2O

Answer 1

Answer on Question #40465 - Chemistry - Other

Question

In the following reaction, how many grams of potassium phosphate, $\mathrm{K}_3\mathrm{PO}_4$, will be produced from $62.3\mathrm{g}$ of potassium hydroxide, KOH?

$$3 \mathrm{KOH} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{K}_3\mathrm{PO}_4 + 3 \mathrm{H}_2\mathrm{O}$$

Answer:

Molar mass of KOH equals:

$$M(\mathrm{KOH}) = M(\mathrm{K}) + M(\mathrm{O}) + M(\mathrm{H}) = 39.1 + 16.0 + 1.008 = 56.108 \frac{\mathrm{g}}{\mathrm{mol}}$$

Mass of 3 moles of potassium hydroxide equals:

$$3 \cdot 56.108 = 168.324 \mathrm{g}$$

Molar mass of $\mathrm{K}_3\mathrm{PO}_4$ equals:

$$M(\mathrm{K}_3\mathrm{PO}_4) = 3M(\mathrm{K}) + M(\mathrm{P}) + 4M(\mathrm{O}) = 3 \cdot 39.1 + 30.97 + 4 \cdot 16.0 = 212.27 \frac{\mathrm{g}}{\mathrm{mol}}$$

Therefore, mass of 1 mole of potassium phosphate equals $212.27\mathrm{g}$.

Then we make a proportion:

- 168.324 g of KOH react to produce 212.27 g of $\mathrm{K}_3\mathrm{PO}_4$

- 62.3 g of KOH - x g of $\mathrm{K}_3\mathrm{PO}_4$

- $x = \frac{62.3 \cdot 212.27}{168.324} = 78.6\mathrm{g}$

Answer: $\mathrm{m}(\mathrm{K}_3\mathrm{PO}_4) = 78.6\mathrm{g}$