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Sagot :
Let's walk through the problem step-by-step to determine how many grams of the solid product, silver chloride (AgCl), will form when 22.0 grams of aluminum chloride (AlCl₃) reacts with 26.0 grams of silver nitrate (AgNO₃).
### Step 1: Molar Mass Calculation
First, we need the molar masses of the reactants and the product involved in this reaction:
- Molar mass of Aluminum chloride (AlCl₃) ≈ 133.34 g/mol
- Molar mass of Silver nitrate (AgNO₃) ≈ 169.87 g/mol
- Molar mass of Silver chloride (AgCl) ≈ 143.32 g/mol
### Step 2: Moles Calculation
Next, we calculate the number of moles of each reactant:
- Moles of AlCl₃:
[tex]\[ \text{Moles of AlCl}_3 = \frac{\text{Mass of AlCl$_3$}}{\text{Molar mass of AlCl$_3$}} = \frac{22.0 \, \text{g}}{133.34 \, \text{g/mol}} \approx 0.165 \, \text{mol} \][/tex]
- Moles of AgNO₃:
[tex]\[ \text{Moles of AgNO}_3 = \frac{\text{Mass of AgNO$_3$}}{\text{Molar mass of AgNO$_3$}} = \frac{26.0 \, \text{g}}{169.87 \, \text{g/mol}} \approx 0.153 \, \text{mol} \][/tex]
### Step 3: Limiting Reagent Determination
The balanced chemical equation is:
[tex]\[ \text{AlCl}_3 + 3\text{AgNO}_3 \rightarrow 3\text{AgCl} + \text{Al(NO}_3\text{)}_3 \][/tex]
According to the stoichiometry of the reaction, 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.
- To react completely with 0.165 moles of AlCl₃, the required moles of AgNO₃:
[tex]\[ \text{Required moles of AgNO}_3 = 0.165 \, \text{mol} \times 3 = 0.495 \, \text{mol} \][/tex]
- We only have 0.153 moles of AgNO₃ available, so AgNO₃ is the limiting reagent.
### Step 4: Product Calculation
Since AgNO₃ is the limiting reagent, the amount of product (AgCl) formed is determined by the moles of AgNO₃:
- The reaction produces 1 mole of AgCl for each mole of AgNO₃, so:
[tex]\[ \text{Moles of AgCl} = 0.153 \, \text{mol} \][/tex]
- To find the mass of AgCl produced:
[tex]\[ \text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl} = 0.153 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 21.936 \, \text{g} \][/tex]
### Conclusion
The mass of the solid product (AgCl) expected to form is approximately 21.9 g.
So, the correct answer is 21.9 g.
### Step 1: Molar Mass Calculation
First, we need the molar masses of the reactants and the product involved in this reaction:
- Molar mass of Aluminum chloride (AlCl₃) ≈ 133.34 g/mol
- Molar mass of Silver nitrate (AgNO₃) ≈ 169.87 g/mol
- Molar mass of Silver chloride (AgCl) ≈ 143.32 g/mol
### Step 2: Moles Calculation
Next, we calculate the number of moles of each reactant:
- Moles of AlCl₃:
[tex]\[ \text{Moles of AlCl}_3 = \frac{\text{Mass of AlCl$_3$}}{\text{Molar mass of AlCl$_3$}} = \frac{22.0 \, \text{g}}{133.34 \, \text{g/mol}} \approx 0.165 \, \text{mol} \][/tex]
- Moles of AgNO₃:
[tex]\[ \text{Moles of AgNO}_3 = \frac{\text{Mass of AgNO$_3$}}{\text{Molar mass of AgNO$_3$}} = \frac{26.0 \, \text{g}}{169.87 \, \text{g/mol}} \approx 0.153 \, \text{mol} \][/tex]
### Step 3: Limiting Reagent Determination
The balanced chemical equation is:
[tex]\[ \text{AlCl}_3 + 3\text{AgNO}_3 \rightarrow 3\text{AgCl} + \text{Al(NO}_3\text{)}_3 \][/tex]
According to the stoichiometry of the reaction, 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.
- To react completely with 0.165 moles of AlCl₃, the required moles of AgNO₃:
[tex]\[ \text{Required moles of AgNO}_3 = 0.165 \, \text{mol} \times 3 = 0.495 \, \text{mol} \][/tex]
- We only have 0.153 moles of AgNO₃ available, so AgNO₃ is the limiting reagent.
### Step 4: Product Calculation
Since AgNO₃ is the limiting reagent, the amount of product (AgCl) formed is determined by the moles of AgNO₃:
- The reaction produces 1 mole of AgCl for each mole of AgNO₃, so:
[tex]\[ \text{Moles of AgCl} = 0.153 \, \text{mol} \][/tex]
- To find the mass of AgCl produced:
[tex]\[ \text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl} = 0.153 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 21.936 \, \text{g} \][/tex]
### Conclusion
The mass of the solid product (AgCl) expected to form is approximately 21.9 g.
So, the correct answer is 21.9 g.
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