- Calculate the moles of barium chloride using the molarity and volume provided: \(n_B = 3.00 \text{ mol/L} \times 0.0200 \text{ L} = 0.0600 \text{ mol}\).
- Determine the moles of aluminum sulfite needed for the reaction using the 1:3 stoichiometric ratio: \(n_A = \frac{0.0600}{3} = 0.0200 \text{ mol}\).
- Calculate the molarity of the aluminum sulfite solution by dividing the moles by its volume: \(M_A = \frac{0.0200 \text{ mol}}{0.0500 \text{ L}} = 0.400 \text{ M}\).
- The final concentration of the aluminum sulfite solution is \(\boxed{0.400 \text{ M}}\).
Explanation
- Problem Analysis and Data Review
We have the balanced chemical equation for the reaction between aluminum sulfite and barium chloride:
\[Al_2(SO_3)_3 + 3BaCl_2 \rightarrow 2AlCl_3 + 3BaSO_3\]
The problem provides the following data:
- Volume of aluminum sulfite solution (\(V_A\)): \(50.0 \text{ mL} = 0.0500 \text{ L}\)
- Volume of barium chloride solution (\(V_B\)): \(20.0 \text{ mL} = 0.0200 \text{ L}\)
- Molarity of barium chloride solution (\(M_B\)): \(3.00 \text{ M}\)
Our goal is to find the molarity of the aluminum sulfite solution (\(M_A\)). We know that the reaction is complete, meaning the stoichiometry is exactly as shown in the equation.
- Calculating Moles of Barium Chloride
First, we calculate the number of moles of barium chloride (\(n_B\)) involved in the reaction using its molarity and volume:
\[n_B = M_B \times V_B = 3.00 \text{ mol/L} \times 0.0200 \text{ L} = 0.0600 \text{ mol}\]
Thus, there are \(0.0600\) moles of \(BaCl_2\) used in the reaction.
- Determining Moles of Aluminum Sulfite
From the balanced chemical equation, we see that 1 mole of \(Al_2(SO_3)_3\) reacts with 3 moles of \(BaCl_2\). This ratio tells us how much aluminum sulfite is required for the reaction:
\[\frac{n_A}{n_B} = \frac{1}{3} \implies n_A = \frac{1}{3} \times n_B\]
Substituting the known value of \(n_B\):
\[n_A = \frac{1}{3} \times 0.0600 \text{ mol} = 0.0200 \text{ mol}\]
So, there were \(0.0200\) moles of aluminum sulfite in the initial solution.
- Calculating Molarity of Aluminum Sulfite
Finally, we determine the molarity of the aluminum sulfite solution (\(M_A\)) by dividing the moles of solute by the volume of the solution in liters:
\[M_A = \frac{n_A}{V_A} = \frac{0.0200 \text{ mol}}{0.0500 \text{ L}} = 0.400 \text{ M}\]
The molarity of the aluminum sulfite solution is \(0.400 \text{ M}\). Therefore, the final answer is \(\boxed{0.400 \text{ M}}\).
Examples
In laboratories, determining the concentration of a chemical reagent is vital for precise synthesis and analysis. For instance, if a researcher needs to react an unknown solution of aluminum sulfite with a known amount of barium chloride to form a precipitate, they can use the stoichiometry of the reaction to deduce the concentration of the unknown solution. This process, often referred to as titration, is a fundamental technique in chemistry that ensures reactions occur with the expected ratios, allowing for accurate and reproducible experimental outcomes.