I was doing a lab experiment on the reaction of copper(II) sulphate and aluminum:
1) I measured 2.02g of copper(II) sulphate pentahydrate
2) I dissolved the copper(II) sulphate pentahydrate in 10mL of distilled water
3) I added 2.0mL of concentrated HCl to the solution and mixed well
4) I added 0.25g on Aluminun foil to the solution.
5) After 5 minutes, i added an additional 5mL of concentrated HCl
6) After all the aluminum foil has reacted, i decanted the solution from the solid (leaving copper behind).
7) The mass of copper product after drying it was 0.69g.
Now we have:
-Mass of CuSO4*5H2O = 2.02g
-Mass of Aluminuim foil = 0.25g
-Mass of copper metal product = 0.69g
The questions are:
1) Calculate moles of Al used.
2) Calculate the moles of CuSO4*5H2O used.
3) Calculate the moles of copper product based on moles of Al.
4) Calculate the moles of copper product based on moles of CuSO4*5H2O.
5) Whats the limiting reactant.
6) Whats the grams of copper product based on the limiting reactant (theoretical yiel).
7) Calculate the percent yield of Copper.
And where are you on this problem; i.e., how much do you know how to do.
To get started, mols Al = grams/atomic mass = 0.25/26.98 = ?
0.00113
I need answer
To answer these questions, we'll need to use mole-to-mole ratios and stoichiometry. Let's go through each question step by step:
1) To calculate the moles of Al used, we need to know the molar mass of Al. Aluminum has an atomic mass of approximately 26.98 g/mol. We can calculate the moles of Al using the following equation:
moles of Al = mass of Al / molar mass of Al
Substituting the values:
moles of Al = 0.25g / 26.98 g/mol
2) To calculate the moles of CuSO4 * 5H2O used, we'll use the molar mass of CuSO4 * 5H2O. Copper sulfate pentahydrate has a molar mass of approximately 249.68 g/mol. We can calculate the moles of CuSO4 * 5H2O using the equation:
moles of CuSO4 * 5H2O = mass of CuSO4 * 5H2O / molar mass of CuSO4 * 5H2O
Substituting the values:
moles of CuSO4 * 5H2O = 2.02g / 249.68 g/mol
3) To calculate the moles of copper product based on moles of Al, we need to know the balanced chemical equation for the reaction. The reaction between Al and CuSO4 is:
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
From the balanced equation, we can see that the stoichiometric ratio between Al and Cu is 2:3. Therefore, for every 2 moles of Al used, we should produce 3 moles of Cu.
moles of Cu product = (moles of Al x 3) / 2
Substituting the previously calculated value of moles of Al:
moles of Cu product = (moles of Al x 3) / 2
4) To calculate the moles of copper product based on moles of CuSO4 * 5H2O, we'll use the same balanced equation. From the equation, we see that the stoichiometric ratio between CuSO4 * 5H2O and Cu is 3:3, or 1:1.
moles of Cu product = moles of CuSO4 * 5H2O
Substituting the previously calculated value of moles of CuSO4 * 5H2O:
moles of Cu product = moles of CuSO4 * 5H2O
5) To determine the limiting reactant, we compare the moles of Cu product based on moles of Al and moles of CuSO4 * 5H2O calculated in questions 3 and 4. The reactant that produces fewer moles of Cu product is the limiting reactant.
6) To calculate the theoretical yield (in grams) of copper product based on the limiting reactant, we'll use the molar mass of Cu. Copper has an atomic mass of approximately 63.55 g/mol. We'll use the following equation:
theoretical yield = moles of Cu product x molar mass of Cu
Substituting the moles of Cu product based on the limiting reactant into the equation:
theoretical yield = moles of Cu product x molar mass of Cu
7) To calculate the percent yield of copper, we need to know the actual yield of copper obtained (0.69g in this case), and then use the following equation:
percent yield = (actual yield / theoretical yield) x 100
Substituting the values:
percent yield = (0.69g / theoretical yield) x 100
This will give us the percentage of the theoretical yield of copper that was obtained in the experiment.