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Multiple Choice
100.0 mL of 0.100 M sodium phosphate is mixed with 100.0 mL of 0.120 M calcium chloride. What mass of calcium phosphate precipitate is formed?
A
0.620 g
B
0.480 g
C
0.240 g
D
0.360 g
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1
Write the balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and calcium chloride (CaCl2). The reaction forms calcium phosphate (Ca3(PO4)2) and sodium chloride (NaCl). The balanced equation is: 2 Na3PO4 + 3 CaCl2 → Ca3(PO4)2 + 6 NaCl.
Calculate the initial moles of sodium phosphate and calcium chloride. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For sodium phosphate: \( 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.010 \, \text{moles} \). For calcium chloride: \( 0.120 \, \text{M} \times 0.100 \, \text{L} = 0.012 \, \text{moles} \).
Determine the limiting reactant by comparing the mole ratio from the balanced equation. According to the equation, 2 moles of Na3PO4 react with 3 moles of CaCl2. Calculate the required moles of each reactant to fully react with the other: \( \frac{0.010 \, \text{moles Na3PO4}}{2} \times 3 = 0.015 \, \text{moles CaCl2} \) needed, but only 0.012 moles are available, so CaCl2 is the limiting reactant.
Calculate the moles of calcium phosphate formed using the stoichiometry of the balanced equation. From the equation, 3 moles of CaCl2 produce 1 mole of Ca3(PO4)2. Therefore, \( \frac{0.012 \, \text{moles CaCl2}}{3} = 0.004 \, \text{moles Ca3(PO4)2} \) are formed.
Convert the moles of calcium phosphate to mass using its molar mass. The molar mass of Ca3(PO4)2 is approximately 310.18 g/mol. Use the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \). Calculate: \( 0.004 \, \text{moles} \times 310.18 \, \text{g/mol} \) to find the mass of the precipitate.