Solution Stoichiometry: Please answer all questions, showing work.ÂÂ
A. What volume of 0.175 M Na3PO4 solution is required to completely react with 126mL of 0.150 M 0.102 M CuCl2?
2 Na3PO4 (aq) + 3 CuCl2 (aq) -> Cu3(PO4)2 (s) + 6 NaCl (aq)
B. What is the Molarity of ZnCl2 that forms when 25.0 g of zinc completely react with a solution of CuCl2? Assume a final volume of 265 mL.ÂÂ
2 Al (s) + 3 H2SO4 (aq) -> Al2(SO4)3 (aq) + 3 H2 (g)
C. What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M sodium carbonate solution?ÂÂ
Na2CO3 (aq) + 2HNO3 (aq) -> 2 NaNO3 (aq) + CO2 (g) + H2O (l)
What mass of carbon dioxide forms?ÂÂ
The Correct Answer and Explanation is:
A. Volume of 0.175 M Na₃PO₄ needed to react with 126 mL of 0.150 M CuCl₂
Given:
- Reaction:
$2 \text{Na}_3\text{PO}_4 (aq) + 3 \text{CuCl}_2 (aq) \to \text{Cu}_3(\text{PO}_4)_2 (s) + 6 \text{NaCl} (aq)$ - Volume of CuCl₂ = 126 mL = 0.126 L
- Molarity of CuCl₂ = 0.150 M
- Molarity of Na₃PO₄ = 0.175 M
Step 1: Calculate moles of CuCl₂
$$
\text{moles CuCl}_2 = M \times V = 0.150 \, \text{mol/L} \times 0.126 \, \text{L} = 0.0189 \, \text{mol}
$$
Step 2: Use mole ratio from balanced equation to find moles of Na₃PO₄ needed:
$$
\frac{2 \, \text{mol Na}_3\text{PO}_4}{3 \, \text{mol CuCl}_2} = x \implies x = 0.0189 \times \frac{2}{3} = 0.0126 \, \text{mol Na}_3\text{PO}_4
$$
Step 3: Calculate volume of 0.175 M Na₃PO₄ solution that contains 0.0126 mol:
$$
V = \frac{\text{moles}}{M} = \frac{0.0126}{0.175} = 0.072 \, \text{L} = 72.0 \, \text{mL}
$$
Answer A: 72.0 mL of 0.175 M Na₃PO₄ solution is required.
B. Molarity of ZnCl₂ formed from 25.0 g zinc reacting with CuCl₂ solution, volume = 265 mL
Note: The reaction for zinc with CuCl₂ was not provided, but typically zinc displaces copper in:
$$
\text{Zn (s)} + \text{CuCl}_2 (aq) \to \text{ZnCl}_2 (aq) + \text{Cu (s)}
$$
Given:
- Mass Zn = 25.0 g
- Volume final solution = 265 mL = 0.265 L
Step 1: Calculate moles of zinc:
Atomic mass Zn = 65.38 g/mol
$$
\text{moles Zn} = \frac{25.0}{65.38} = 0.3823 \, \text{mol}
$$
Step 2: From the reaction, 1 mole Zn produces 1 mole ZnCl₂, so:
$$
\text{moles ZnCl}_2 = 0.3823 \, \text{mol}
$$
Step 3: Calculate molarity of ZnCl₂:
$$
M = \frac{\text{moles}}{\text{volume in liters}} = \frac{0.3823}{0.265} = 1.44 \, \text{M}
$$
Answer B: The molarity of ZnCl₂ formed is 1.44 M.
C. Volume of 0.150 M HNO₃ required to react with 35.7 mL of 0.108 M Na₂CO₃, and mass of CO₂ formed
Reaction:
$$
\text{Na}_2\text{CO}_3 (aq) + 2 \text{HNO}_3 (aq) \to 2 \text{NaNO}_3 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l)
$$
Given:
- Volume Na₂CO₃ = 35.7 mL = 0.0357 L
- Molarity Na₂CO₃ = 0.108 M
- Molarity HNO₃ = 0.150 M
Step 1: Calculate moles Na₂CO₃:
$$
\text{moles Na}_2\text{CO}_3 = 0.108 \times 0.0357 = 0.003856 \, \text{mol}
$$
Step 2: Use mole ratio for HNO₃ to Na₂CO₃ (2:1):
$$
\text{moles HNO}_3 = 0.003856 \times 2 = 0.007712 \, \text{mol}
$$
Step 3: Calculate volume of 0.150 M HNO₃ needed:
$$
V = \frac{\text{moles}}{M} = \frac{0.007712}{0.150} = 0.05141 \, \text{L} = 51.4 \, \text{mL}
$$
Mass of CO₂ formed
From reaction, 1 mole Na₂CO₃ produces 1 mole CO₂.
Step 4: Moles CO₂ = moles Na₂CO₃ = 0.003856 mol
Step 5: Calculate mass CO₂:
Molar mass CO₂ = 44.01 g/mol
$$
\text{mass CO}_2 = 0.003856 \times 44.01 = 0.1697 \, \text{g}
$$
Answer C:
- Volume of HNO₃ required = 51.4 mL
- Mass of CO₂ formed = 0.170 g
Summary:
| Part | Answer |
|---|---|
| A | 72.0 mL of 0.175 M Na₃PO₄ |
| B | 1.44 M ZnCl₂ solution formed |
| C | 51.4 mL of 0.150 M HNO₃; 0.170 g CO₂ formed |
Stoichiometry allows us to calculate the amounts of reactants and products in chemical reactions by using balanced chemical equations. Each equation provides the mole ratios between substances, which is key to solving these problems.
In part A, the balanced equation $2 \text{Na}_3\text{PO}_4 + 3 \text{CuCl}_2 \to \text{Cu}_3(\text{PO}_4)_2 + 6 \text{NaCl}$ tells us that 2 moles of sodium phosphate react with 3 moles of copper chloride. Given the molarity and volume of CuCl₂, we find moles of CuCl₂ and then calculate the corresponding moles of Na₃PO₄ using the mole ratio. Finally, converting moles of Na₃PO₄ to volume using its molarity yields the volume required.
Part B involves a displacement reaction where zinc metal reacts with CuCl₂ solution forming ZnCl₂ and copper metal. Since 1 mole of zinc produces 1 mole of ZnCl₂, the moles of zinc directly give moles of ZnCl₂. Using the final solution volume, we compute molarity as moles/volume. This approach applies conservation of mass and mole equivalency in reactions.
Part C requires reacting sodium carbonate with nitric acid. The balanced equation shows that 2 moles of HNO₃ react with 1 mole of Na₂CO₃. Calculating moles of Na₂CO₃ from given concentration and volume, then doubling it to find moles of HNO₃ needed, allows us to find the volume of HNO₃ required. Additionally, 1 mole of Na₂CO₃ produces 1 mole of CO₂ gas. Converting moles of CO₂ to grams using molar mass gives the mass of carbon dioxide formed.
Understanding mole ratios, molarity, and volume relationships is critical in stoichiometry. These calculations are foundational in chemistry, especially for preparing solutions, reacting chemicals quantitatively, and predicting yields.
