Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

6. Chemical Quantities & Aqueous Reactions

Solution Stoichiometry

General Chemistry

6. Chemical Quantities & Aqueous Reactions

Solution Stoichiometry

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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?

0.620 g

0.480 g

0.240 g

0.360 g

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Verified step by step guidance

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.