Question

bien sûr ! essaie celle-ci tout seul :

exercice

\bacl_2(aq) + na_2so_4(aq) \
ightarrow baso_4(s) + 2nacl(aq)\\

on mélange :

• \\(150\text{ ml}\\) de \\(bacl_2\\) à \\(0{,}80\text{ mol/l}\\)
• \\(250\text{ ml}\\) de \\(na_2so_4\\) à \\(0{,}60\text{ mol/l}\\)

trouve :

1. le réactif limitant.
2. la masse de \\(baso_4\\) produite.
3. la concentration finale de \\(nacl\\).

Explanation:

🆕 New Concept Discovered: Limiting Reactant and Solution Stoichiometry
Finding which reactant runs out first to calculate product yields in solution

Step 1: Calculate initial moles of reactants

Using the formula \( n = C \times V \):

For \(\text{BaCl}_2\):

$$ V = 150\text{ mL} = 0.150\text{ L} $$

$$ n(\text{BaCl}_2) = 0.80\text{ mol/L} \times 0.150\text{ L} = 0.12\text{ mol} $$

For \(\text{Na}_2\text{SO}_4\):

$$ V = 250\text{ mL} = 0.250\text{ L} $$

$$ n(\text{Na}_2\text{SO}_4) = 0.60\text{ mol/L} \times 0.250\text{ L} = 0.15\text{ mol} $$

Step 2: Determine the limiting reactant

The balanced chemical equation is:

$$ \text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) ightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq) $$

The stoichiometric ratio between \(\text{BaCl}_2\) and \(\text{Na}_2\text{SO}_4\) is \(1:1\).

Comparing the available moles:

• We have \(0.12\text{ mol}\) of \(\text{BaCl}_2\).
• We have \(0.15\text{ mol}\) of \(\text{Na}_2\text{SO}_4\).

Since \(0.12\text{ mol} < 0.15\text{ mol}\), \(\text{BaCl}_2\) will be completely consumed first.

• Limiting reactant: \(\text{BaCl}_2\)

Step 3: Calculate the mass of \(\text{BaSO}_4\) produced

The mole ratio between the limiting reactant \(\text{BaCl}_2\) and \(\text{BaSO}_4\) is \(1:1\).

$$ n(\text{BaSO}_4) = n(\text{BaCl}_2) = 0.12\text{ mol} $$

Calculate the molar mass of \(\text{BaSO}_4\):

$$ M(\text{BaSO}_4) = M(\text{Ba}) + M(\text{S}) + 4 \times M(\text{O}) $$

$$ M(\text{BaSO}_4) = 137.33 + 32.06 + 4 \times 16.00 = 233.39\text{ g/mol} $$

Calculate the mass:

$$ m(\text{BaSO}_4) = n \times M = 0.12\text{ mol} \times 233.39\text{ g/mol} \approx 28.01\text{ g} $$

Step 4: Calculate the final concentration of \(\text{NaCl}\)

The mole ratio between \(\text{BaCl}_2\) and \(\text{NaCl}\) is \(1:2\).

$$ n(\text{NaCl}) = 2 \times n(\text{BaCl}_2) = 2 \times 0.12\text{ mol} = 0.24\text{ mol} $$

The total volume of the mixture is:

$$ V_{\text{total}} = 150\text{ mL} + 250\text{ mL} = 400\text{ mL} = 0.400\text{ L} $$

Calculate the final concentration of \(\text{NaCl}\):

$$ C(\text{NaCl}) = \frac{n(\text{NaCl})}{V_{\text{total}}} = \frac{0.24\text{ mol}}{0.400\text{ L}} = 0.60\text{ mol/L} $$

Answer:

1. Le réactif limitant : \(\text{BaCl}_2\)
2. La masse de \(\text{BaSO}_4\) produite : \(28.01\text{ g}\) (or \(28\text{ g}\) using two significant figures)
3. La concentration finale de \(\text{NaCl}\) : \(0.60\text{ mol/L}\)