Question

exercice

\\(agno_3(aq) + nacl(aq) \
ightarrow agcl(s) + nano_3(aq)\\)

on mélange :

• \\(250\text{ ml}\\) de \\(agno_3\\) à \\(0,40\text{ mol/l}\\)
• \\(300\text{ ml}\\) de \\(nacl\\) à \\(0,60\text{ mol/l}\\)

trouve :

1. le réactif limitant.
2. la masse de \\(agcl\\) produite.
3. la concentration finale de \\(nano_3\\).
4. combien de moles du réactif en excès restent après la réaction (comme le \\(-0,10\\) de ton prof).

Explanation:

🆕 New Concept Discovered: Limiting Reactant and Excess Calculations
Finding which reactant runs out first and what remains

Step 1: Calculate initial moles of reactants

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

For \( \text{AgNO}_3 \):

$$ V_1 = 250\text{ mL} = 0.250\text{ L} $$

$$ C_1 = 0.40\text{ mol/L} $$

$$ n(\text{AgNO}_3) = 0.40\text{ mol/L} \times 0.250\text{ L} = 0.10\text{ mol} $$

For \( \text{NaCl} \):

$$ V_2 = 300\text{ mL} = 0.300\text{ L} $$

$$ C_2 = 0.60\text{ mol/L} $$

$$ n(\text{NaCl}) = 0.60\text{ mol/L} \times 0.300\text{ L} = 0.18\text{ mol} $$

Step 2: Determine the limiting reactant

The balanced chemical equation is:

$$ \text{AgNO}_3(aq) + \text{NaCl}(aq) ightarrow \text{AgCl}(s) + \text{NaNO}_3(aq) $$

The stoichiometric ratio is \( 1:1 \).

• Since we have \( 0.10\text{ mol} \) of \( \text{AgNO}_3 \) and \( 0.18\text{ mol} \) of \( \text{NaCl} \), \( \text{AgNO}_3 \) will be completely consumed first.
• Limiting reactant: \( \text{AgNO}_3 \)
• Excess reactant: \( \text{NaCl} \)

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

Since \( \text{AgNO}_3 \) is the limiting reactant, it determines the amount of product formed:

$$ n(\text{AgCl}) = n(\text{AgNO}_3) = 0.10\text{ mol} $$

Molar mass of \( \text{AgCl} \):

$$ M(\text{AgCl}) = 107.87\text{ g/mol (Ag)} + 35.45\text{ g/mol (Cl)} = 143.32\text{ g/mol} $$

Mass of \( \text{AgCl} \):

$$ m(\text{AgCl}) = n \times M = 0.10\text{ mol} \times 143.32\text{ g/mol} = 14.33\text{ g} $$

Step 4: Calculate the final concentration of \( \text{NaNO}_3 \)

The moles of \( \text{NaNO}_3 \) produced are equal to the moles of the limiting reactant consumed:

$$ n(\text{NaNO}_3) = 0.10\text{ mol} $$

The total volume of the mixture is:

$$ V_{\text{total}} = 250\text{ mL} + 300\text{ mL} = 550\text{ mL} = 0.550\text{ L} $$

Final concentration:

$$ C(\text{NaNO}_3) = \frac{n}{V_{\text{total}}} = \frac{0.10\text{ mol}}{0.550\text{ L}} \approx 0.18\text{ mol/L} $$

Step 5: Calculate remaining moles of excess reactant

The moles of \( \text{NaCl} \) consumed during the reaction:

$$ n(\text{NaCl})_{\text{consumed}} = n(\text{AgNO}_3) = 0.10\text{ mol} $$

The remaining moles of \( \text{NaCl} \) are:

$$ n(\text{NaCl})_{\text{remaining}} = n(\text{NaCl})_{\text{initial}} - n(\text{NaCl})_{\text{consumed}} $$

$$ n(\text{NaCl})_{\text{remaining}} = 0.18\text{ mol} - 0.10\text{ mol} = 0.08\text{ mol} $$

Answer:

1. Réactif limitant : \( \text{AgNO}_3 \)
2. Masse de \( \text{AgCl} \) produite : \( 14.33\text{ g} \)
3. Concentration finale de \( \text{NaNO}_3 \) : \( 0.18\text{ mol/L} \)
4. Moles de réactif en excès restantes : \( 0.08\text{ mol} \) de \( \text{NaCl} \)