Respuesta :
Answer:
A) Solid copper, solid aluminum, and aqueous aluminum chloride
Explanation:
Let's consider the following reaction.
2 Al(s) + 3 CuCl₂(aq) → 2 AlCl₃(aq) + 3 Cu
The molar mass of Al is 26.98 g/mol. The moles of Al are:
0.786 g × (1 mol / 26.98 g) = 0.0291 mol
The moles of CuCl₂ are:
25.0 × 10⁻³ L × 0.21 mol/L = 0.0053 mol
The theoretical molar ratio of Al to CuCl₂ is 2/3 = 0.66/1
The experimental molar ratio of Al to CuCl₂ is 0.0291 mol / 0.0053 mol = 5.49/1.
Comparing both molar ratios, we can determine that the limiting reactant is CuCl₂, that is, CuCl₂ will consume completely.
After the reaction is completed, there will be Al, Cu and AlCl₃.
Answer:
In the flask solid copper, solid aluminum, and aqueous aluminum chloride will be present after the reaction
Explanation:
Step 1: Data given
Mass of aluminium = 0.786 grams
Volume of 0.21 M CuCl2 solution = 25.0 mL = 0.025 L
Molar mass of Al = 26.98 g/mol
Step 2: The balanced equation
CuCl2 + Al → AlCl2 + Cu
Step 3: Calculate moles Al
Moles Al = mass Al / molar mass Al
Moles Al = 0.786 grams / 26.98 g/mol
Moles Al = 0.0291 moles
Step 4: Calculate moles CuCl2
Moles CuCl2 = molarity * volume
Moles CuCl2 = 0.21 M * 0.025 L
Moles CuCl2 = 0.00525 moles
Step 5: Calculate the limiting reactant
For 1 mol CuCl2 we need 1 mol Al to produce 1 mol AlCl2 and 1 mol Cu
CuCl2 is the limiting reactant, it will completely be consumed ( 0.00525 moles).
Al is in excess. There will react 0.00525 moles. There will remain 0.0291 - 0.00525 = 0.02385 moles Al
Step 6: Calculate moles of products
There will be produce 0.00525 moles of AlCl2 and 0.00525 moles of Cu
This means in the flask solid copper, solid aluminum, and aqueous aluminum chloride will be present after the reaction
