Domanda aperta
Consider the precipitation reaction: 2 Na3PO4(aq) + 3 CuCl2(aq) → Cu3(PO4)2(s) + 6 NaCl(aq). What volume of 0.175 M Na3PO4 solution is necessary to completely react with 95.4 mL of 0.102 M CuCl2?
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6. Chemical Quantities & Aqueous Reactions
Solution Stoichiometry
Scelta multipla
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
Guida verificata passo dopo passo1
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.
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