Answer :
To determine the limiting reactant, we need to follow these steps:
1. Calculate the molar masses of the reactants:
- Molar mass of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Cu} = 63.55 \, \text{g/mol}, \quad \text{Cl} = 35.45 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \, \text{CuCl}_2 = 63.55 \, \text{g/mol} + 2 \times 35.45 \, \text{g/mol} = 134.45 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Al} = 26.98 \, \text{g/mol} \][/tex]
2. Determine the given masses of the reactants:
- Mass of [tex]\( \text{CuCl}_2 \)[/tex] is [tex]\( 10.5 \, \text{g} \)[/tex]
- Mass of [tex]\( \text{Al} \)[/tex] is [tex]\( 12.4 \, \text{g} \)[/tex]
3. Calculate the number of moles of each reactant:
- Moles of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Moles of} \, \text{CuCl}_2 = \frac{10.5 \, \text{g}}{134.45 \, \text{g/mol}} \approx 0.0781 \, \text{moles} \][/tex]
- Moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Moles of} \, \text{Al} = \frac{12.4 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.4596 \, \text{moles} \][/tex]
4. Use the stoichiometry of the balanced chemical equation to determine the mole ratios and the requirements for the reaction:
The balanced equation is:
[tex]\[ 3 \, \text{CuCl}_2 (aq) + 2 \, \text{Al} (s) \rightarrow 3 \, \text{Cu} (s) + 2 \, \text{AlCl}_3 (aq) \][/tex]
- For the given moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Moles of} \, \text{CuCl}_2 \, \text{needed} = \left( \frac{3}{2} \right) \times \text{Moles of} \, \text{Al} \][/tex]
[tex]\[ \text{Moles of} \, \text{CuCl}_2 \, \text{needed} = \left( \frac{3}{2} \right) \times 0.4596 \, \text{moles} \approx 0.6894 \, \text{moles} \][/tex]
- For the given moles of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Moles of} \, \text{Al} \, \text{needed} = \left( \frac{2}{3} \right) \times \text{Moles of} \, \text{CuCl}_2 \][/tex]
[tex]\[ \text{Moles of} \, \text{Al} \, \text{needed} = \left( \frac{2}{3} \right) \times 0.0781 \, \text{moles} \approx 0.0521 \, \text{moles} \][/tex]
5. Determine the limiting reactant by comparing the calculated moles needed to the moles available:
- Available moles of [tex]\( \text{CuCl}_2 \)[/tex]: [tex]\( 0.0781 \, \text{moles} \)[/tex]
- Moles of [tex]\( \text{CuCl}_2 \, \text{needed for given} \, \text{Al} \)[/tex]: [tex]\( 0.6894 \, \text{moles} \)[/tex]
Since [tex]\( 0.0781 \, \text{moles} \)[/tex] of [tex]\( \text{CuCl}_2 \)[/tex] is much less than the [tex]\( 0.6894 \, \text{moles} \)[/tex] needed, [tex]\( \text{CuCl}_2 \)[/tex] is the limiting reactant.
Therefore, the limiting reactant in this chemical reaction is:
[tex]\[ \boxed{\text{CuCl}_2} \][/tex]
1. Calculate the molar masses of the reactants:
- Molar mass of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Cu} = 63.55 \, \text{g/mol}, \quad \text{Cl} = 35.45 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \, \text{CuCl}_2 = 63.55 \, \text{g/mol} + 2 \times 35.45 \, \text{g/mol} = 134.45 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Al} = 26.98 \, \text{g/mol} \][/tex]
2. Determine the given masses of the reactants:
- Mass of [tex]\( \text{CuCl}_2 \)[/tex] is [tex]\( 10.5 \, \text{g} \)[/tex]
- Mass of [tex]\( \text{Al} \)[/tex] is [tex]\( 12.4 \, \text{g} \)[/tex]
3. Calculate the number of moles of each reactant:
- Moles of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Moles of} \, \text{CuCl}_2 = \frac{10.5 \, \text{g}}{134.45 \, \text{g/mol}} \approx 0.0781 \, \text{moles} \][/tex]
- Moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Moles of} \, \text{Al} = \frac{12.4 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.4596 \, \text{moles} \][/tex]
4. Use the stoichiometry of the balanced chemical equation to determine the mole ratios and the requirements for the reaction:
The balanced equation is:
[tex]\[ 3 \, \text{CuCl}_2 (aq) + 2 \, \text{Al} (s) \rightarrow 3 \, \text{Cu} (s) + 2 \, \text{AlCl}_3 (aq) \][/tex]
- For the given moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Moles of} \, \text{CuCl}_2 \, \text{needed} = \left( \frac{3}{2} \right) \times \text{Moles of} \, \text{Al} \][/tex]
[tex]\[ \text{Moles of} \, \text{CuCl}_2 \, \text{needed} = \left( \frac{3}{2} \right) \times 0.4596 \, \text{moles} \approx 0.6894 \, \text{moles} \][/tex]
- For the given moles of [tex]\( \text{CuCl}_2 \)[/tex]:
[tex]\[ \text{Moles of} \, \text{Al} \, \text{needed} = \left( \frac{2}{3} \right) \times \text{Moles of} \, \text{CuCl}_2 \][/tex]
[tex]\[ \text{Moles of} \, \text{Al} \, \text{needed} = \left( \frac{2}{3} \right) \times 0.0781 \, \text{moles} \approx 0.0521 \, \text{moles} \][/tex]
5. Determine the limiting reactant by comparing the calculated moles needed to the moles available:
- Available moles of [tex]\( \text{CuCl}_2 \)[/tex]: [tex]\( 0.0781 \, \text{moles} \)[/tex]
- Moles of [tex]\( \text{CuCl}_2 \, \text{needed for given} \, \text{Al} \)[/tex]: [tex]\( 0.6894 \, \text{moles} \)[/tex]
Since [tex]\( 0.0781 \, \text{moles} \)[/tex] of [tex]\( \text{CuCl}_2 \)[/tex] is much less than the [tex]\( 0.6894 \, \text{moles} \)[/tex] needed, [tex]\( \text{CuCl}_2 \)[/tex] is the limiting reactant.
Therefore, the limiting reactant in this chemical reaction is:
[tex]\[ \boxed{\text{CuCl}_2} \][/tex]
