Question

what mass of sodium phosphate is required to generate 10.0 grams of aluminum phosphate? molar mass alpo₄ = 121.95 g/mol molar mass na₃po₄ = 163.94 g/mol 2na₃po₄ + al₂(so₄)₃ → 2alpo₄(s) + 3na₂so₄ ? g na₃po₄

Explanation:

Step1: Calculate moles of \( \text{AlPO}_4 \)

Moles of \( \text{AlPO}_4 = \frac{\text{mass of } \text{AlPO}_4}{\text{molar mass of } \text{AlPO}_4} \)
\( \text{Moles of } \text{AlPO}_4 = \frac{10.0\ \text{g}}{121.95\ \text{g/mol}} \approx 0.0820\ \text{mol} \)

Step2: Relate moles of \( \text{AlPO}_4 \) to \( \text{Na}_3\text{PO}_4 \) using stoichiometry

From the balanced equation: \( 2\ \text{mol}\ \text{Na}_3\text{PO}_4
ightarrow 2\ \text{mol}\ \text{AlPO}_4 \), so the mole ratio of \( \text{Na}_3\text{PO}_4 \) to \( \text{AlPO}_4 \) is \( 1:1 \). Thus, moles of \( \text{Na}_3\text{PO}_4 = \) moles of \( \text{AlPO}_4 = 0.0820\ \text{mol} \)

Step3: Calculate mass of \( \text{Na}_3\text{PO}_4 \)

Mass of \( \text{Na}_3\text{PO}_4 = \text{moles of } \text{Na}_3\text{PO}_4 \times \text{molar mass of } \text{Na}_3\text{PO}_4 \)
\( \text{Mass of } \text{Na}_3\text{PO}_4 = 0.0820\ \text{mol} \times 163.94\ \text{g/mol} \approx 13.4\ \text{g} \)

Answer:

\( 13.4 \) (rounded to three significant figures, or more precisely, the calculation gives approximately \( 13.4\ \text{g} \))