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 find the starting mass of each reactant in the reaction, we can use stoichiometry and the given information. The balanced equation shows that 3 moles of KOH react with 1 mole of H3PO4 to produce 1 mole of K3PO4 and 3 moles of water. First, we need to determine how many moles of K3PO4 were produced: - The molar mass of K3PO4 is approximately \( 212.3 \text{ g/mol} \). - Moles of K3PO4 = \( \frac{55.7 \text{ g}}{212.3 \text{ g/mol}} \approx 0.262 \text{ mol} \). From the stoichiometry of the balanced equation: - To produce 0.262 mol of K3PO4, we need: - 0.262 mol \(\times 3 \) = 0.786 mol of KOH. - 0.262 mol of H3PO4. Next, we calculate the mass of KOH and H3PO4: - Molar mass of KOH is approximately \( 56.1 \text{ g/mol} \): - Mass of KOH = \( 0.786 \text{ mol} \times 56.1 \text{ g/mol} \approx 44.1 \text{ g} \). - Molar mass of H3PO4 is approximately \( 98.0 \text{ g/mol} \): - Moles of H3PO4 needed = \( 0.262 \text{ mol} \) (from earlier). - Mass of H3PO4 = \( 0.262 \text{ mol} \times 98.0 \text{ g/mol} \approx 25.7 \text{ g} \). However, it is stated that \( 89.8 \text{ g} \) of H3PO4 remains unreacted. Thus, the starting mass of H3PO4 must be the sum of the reacted amount and the remaining amount: - Total mass of H3PO4 = \( 25.7 \text{ g} + 89.8 \text{ g} = 115.5 \text{ g} \). In summary: - Starting mass of KOH = \( 44.1 \text{ g} \). - Starting mass of H3PO4 = \( 115.5 \text{ g} \).