Question

for the chemical reaction, 3koh + h₃po₄ → k₃po₄ + 3h₂o. what mass of phosphoric acid in grams is needed to produce 2.13 mol of water?

Explanation:

Step1: Determine mole - ratio

From the balanced chemical equation \(3KOH + H_{3}PO_{4}
ightarrow K_{3}PO_{4}+3H_{2}O\), the mole - ratio of \(H_{3}PO_{4}\) to \(H_{2}O\) is \(1:3\).

Step2: Calculate moles of \(H_{3}PO_{4}\)

If \(n(H_{2}O) = 2.13\ mol\), then \(n(H_{3}PO_{4})=\frac{1}{3}\times n(H_{2}O)\). So \(n(H_{3}PO_{4})=\frac{1}{3}\times2.13\ mol = 0.71\ mol\).

Step3: Calculate molar mass of \(H_{3}PO_{4}\)

The molar mass of \(H_{3}PO_{4}\) is \(M=(3\times1.01 + 30.97+4\times16.00)\ g/mol=97.99\ g/mol\).

Step4: Calculate mass of \(H_{3}PO_{4}\)

Using the formula \(m = n\times M\), we have \(m(H_{3}PO_{4})=n(H_{3}PO_{4})\times M(H_{3}PO_{4})\). Substituting the values, \(m(H_{3}PO_{4})=0.71\ mol\times97.99\ g/mol\approx69.6\ g\).

Answer:

69.6 g