Practice Problem BUILD Potassium hydroxide and phosphoric acid react to form potassium phosphate and water according to the equation \( 3 \mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \). Determine the starting mass of each reactant if \( 55.7 \mathrm{~g} \mathrm{~K}_{3} \mathrm{PO}_{4} \) are produced and \( 89.8 \mathrm{~g} \mathrm{H}_{3} \mathrm{PO}_{4} \) remain unreacted.

To determine the starting masses of potassium hydroxide (KOH) and phosphoric acid (H₃PO₄), we'll analyze the given reaction and the provided data. ### **Given Reaction:** \[ 3 \mathrm{KOH}(aq) + \mathrm{H}_3\mathrm{PO}_4(aq) \longrightarrow \mathrm{K}_3\mathrm{PO}_4(aq) + 3 \mathrm{H}_2\mathrm{O}(l) \] ### **Provided Data:** - **Mass of potassium phosphate (K₃PO₄) produced:** 55.7 g - **Mass of phosphoric acid (H₃PO₄) remaining unreacted:** 89.8 g ### **Step-by-Step Solution:** 1. **Determine Molar Masses:** - **KOH:** \( \mathrm{K} + \mathrm{O} + \mathrm{H} = 39.10 + 16.00 + 1.008 = 56.108 \, \text{g/mol} \) - **H₃PO₄:** \( 3\mathrm{H} + \mathrm{P} + 4\mathrm{O} = 3(1.008) + 30.97 + 4(16.00) = 98.00 \, \text{g/mol} \) - **K₃PO₄:** \( 3\mathrm{K} + \mathrm{P} + 4\mathrm{O} = 3(39.10) + 30.97 + 4(16.00) = 212.27 \, \text{g/mol} \) 2. **Calculate Moles of K₃PO₄ Produced:** \[ \text{Moles of } \mathrm{K}_3\mathrm{PO}_4 = \frac{55.7 \, \text{g}}{212.27 \, \text{g/mol}} \approx 0.262 \, \text{mol} \] 3. **Determine Moles of H₃PO₄ Consumed:** The stoichiometry of the reaction shows that 1 mole of H₃PO₄ produces 1 mole of K₃PO₄. \[ \text{Moles of H}_3\mathrm{PO}_4 \text{ consumed} = 0.262 \, \text{mol} \] 4. **Calculate Initial Moles of H₃PO₄:** \[ \text{Remaining moles of H}_3\mathrm{PO}_4 = \frac{89.8 \, \text{g}}{98.00 \, \text{g/mol}} \approx 0.916 \, \text{mol} \] \[ \text{Initial moles of H}_3\mathrm{PO}_4 = 0.262 + 0.916 = 1.178 \, \text{mol} \] 5. **Determine Initial Mass of H₃PO₄:** \[ \text{Mass of H}_3\mathrm{PO}_4 = 1.178 \, \text{mol} \times 98.00 \, \text{g/mol} \approx 115.6 \, \text{g} \] 6. **Calculate Initial Moles of KOH:** From the reaction stoichiometry, 3 moles of KOH react with 1 mole of H₃PO₄. \[ \text{Moles of KOH} = 3 \times 0.262 = 0.786 \, \text{mol} \] 7. **Determine Initial Mass of KOH:** \[ \text{Mass of KOH} = 0.786 \, \text{mol} \times 56.108 \, \text{g/mol} \approx 44.2 \, \text{g} \] ### **Final Answer:** - **Starting mass of KOH:** **44.2 g** - **Starting mass of H₃PO₄:** **115.6 g**