IDNLearn.com makes it easy to get reliable answers from knowledgeable individuals. Find in-depth and trustworthy answers to all your questions from our experienced community members.
Sagot :
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]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
