A precipitation reaction occurs when 50.0 mL of 0.50 M BaCl2(aq) is mixed with 75.0 mL of 0.75 M Na2CO3(aq). The only precipitate is the BaCO3(s) formed. Which is the limiting reactant?
The correct answer and explanation is:
Let’s analyze the precipitation reaction step-by-step.
Given:
- Volume of BaCl₂ = 50.0 mL = 0.0500 L
- Molarity of BaCl₂ = 0.50 M
- Volume of Na₂CO₃ = 75.0 mL = 0.0750 L
- Molarity of Na₂CO₃ = 0.75 M
Step 1: Write the balanced chemical equation
The reaction between barium chloride and sodium carbonate produces barium carbonate precipitate and sodium chloride in solution: BaCl2(aq)+Na2CO3(aq)→BaCO3(s)+2NaCl(aq)\text{BaCl}_2(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow \text{BaCO}_3(s) + 2\text{NaCl}(aq)
Step 2: Calculate moles of each reactant
- Moles of BaCl₂:
moles BaCl2=M×V=0.50 mol/L×0.0500 L=0.025 mol\text{moles BaCl}_2 = M \times V = 0.50 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.025 \, \text{mol}
- Moles of Na₂CO₃:
moles Na2CO3=0.75 mol/L×0.0750 L=0.05625 mol\text{moles Na}_2\text{CO}_3 = 0.75 \, \text{mol/L} \times 0.0750 \, \text{L} = 0.05625 \, \text{mol}
Step 3: Determine the limiting reactant
From the balanced equation:
- 1 mole of BaCl₂ reacts with 1 mole of Na₂CO₃ to produce 1 mole of BaCO₃.
So, the mole ratio is 1:1.
Compare moles available:
- BaCl₂: 0.025 mol
- Na₂CO₃: 0.05625 mol
Since BaCl₂ has fewer moles, it will run out first.
Therefore, BaCl₂ is the limiting reactant.
Explanation (300 words):
In a precipitation reaction involving BaCl₂ and Na₂CO₃, the formation of the solid BaCO₃ depends on the availability of the reactants. To identify the limiting reactant—the substance that is completely consumed first and thus limits the amount of product formed—we first write the balanced chemical equation: BaCl2+Na2CO3→BaCO3(s)+2NaCl\text{BaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{BaCO}_3(s) + 2\text{NaCl}
The stoichiometry indicates that one mole of barium chloride reacts with one mole of sodium carbonate to produce one mole of barium carbonate precipitate.
Next, we calculate the number of moles of each reactant:
- For BaCl₂, with 0.50 M concentration and 50.0 mL volume, the moles are:
0.50 mol/L×0.0500 L=0.025 mol0.50 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.025 \, \text{mol}
- For Na₂CO₃, with 0.75 M concentration and 75.0 mL volume, the moles are:
0.75 mol/L×0.0750 L=0.05625 mol0.75 \, \text{mol/L} \times 0.0750 \, \text{L} = 0.05625 \, \text{mol}
Since the reaction requires a 1:1 mole ratio, we compare the available moles of each reactant. BaCl₂ has 0.025 moles available, whereas Na₂CO₃ has more, at 0.05625 moles. This means BaCl₂ will be consumed first during the reaction because there is less of it in terms of moles.
Hence, BaCl₂ is the limiting reactant. Once all BaCl₂ is used up, the reaction will stop, regardless of the remaining Na₂CO₃. The amount of BaCO₃ precipitate formed will correspond to the amount of BaCl₂ initially present.
Understanding the limiting reactant helps in predicting the theoretical yield of a product and ensures efficient use of chemicals in laboratory and industrial settings.