3CuCl2 *2H2O + 2Al = 3Cu + 2AlCl3 + 6H2O
Determine the number of grams that should have been produced when 1.25g of copper (2) chloride dehydrate reacts with excess aluminum
so i'm not really sure how you're supposed to calculate the number of grams with hydrates??
MCuCl2= 63.55 g/mol + 2(35.45 g/mol) = 134.45 g/mol
MH2O= 2(1.01 g/mol) + 16 g/mol = 18.02 g/mol
134.45 + 18.02 = 152.47? (is this even what im supposed to do?)
mol CuCl2 *H2O= 1.25g x (1 mol/152.47g)
= 0.008198 mol CuCl2 *H2O
mole Cu= 0.008198 mol CuCl2 *H2O x (3 mol Cu/5 mol CuCl2 *H2O)
= 0.004918 mol Cu
MCu= 63.55 g/mol
mass Cu= 0.004918 mol Cu x (63.55g/1 mol)
= 0.3126 g Cu
i'd really appreciate the help thank you for your time
Wow, thank you so much that was very descriptive, I really appreciate it!
To calculate the number of grams of copper produced when 1.25g of copper(II) chloride dihydrate reacts with excess aluminum, you can follow these steps:
Step 1: Calculate the molar mass of CuCl2 * 2H2O
The molar mass of CuCl2 is 63.55 g/mol, and the molar mass of H2O is 18.02 g/mol. Therefore, the molar mass of CuCl2 * 2H2O is:
63.55 g/mol + 2 * (18.02 g/mol) = 99.59 g/mol
Step 2: Convert the given mass of CuCl2 * 2H2O to moles
Use the molar mass calculated in Step 1 to convert the given mass to moles:
1.25 g CuCl2 * 2H2O * (1 mol / 99.59 g) = 0.01254 mol CuCl2 * 2H2O
Step 3: Determine the number of moles of Cu produced
From the balanced chemical equation, we can see that the molar ratio between CuCl2 * 2H2O and Cu is 5:3. Therefore, the number of moles of Cu produced is:
0.01254 mol CuCl2 * 2H2O * (3 mol Cu / 5 mol CuCl2 * 2H2O) = 0.00752 mol Cu
Step 4: Convert the moles of Cu to grams
Finally, use the molar mass of Cu (63.55 g/mol) to convert moles to grams:
0.00752 mol Cu * (63.55 g / 1 mol) = 0.477 g Cu
Therefore, the number of grams of copper produced when 1.25g of copper(II) chloride dihydrate reacts with excess aluminum is approximately 0.477 grams.
To determine the number of grams that should have been produced when 1.25g of copper(II) chloride dihydrate reacts with excess aluminum, you need to follow these steps:
1. Calculate the molar mass of copper(II) chloride dihydrate (CuCl2 * 2H2O):
- The molar mass of Cu is 63.55 g/mol.
- The molar mass of Cl is 35.45 g/mol.
- The molar mass of H2O is 2(1.01 g/mol) + 16 g/mol = 18.02 g/mol.
- So, the molar mass of CuCl2 * 2H2O is:
MCuCl2 * 2H2O = (63.55 g/mol) + 2(35.45 g/mol) + 2(18.02 g/mol) = 134.45 g/mol.
2. Calculate the number of moles of CuCl2 * 2H2O using the given mass:
- Given: 1.25 g of CuCl2 * 2H2O.
- Convert mass to moles using the molar mass:
Moles = Mass / Molar mass = 1.25 g / 134.45 g/mol = 0.009302 mol.
3. Convert the moles of CuCl2 * 2H2O to moles of Cu:
- To do this, use the balanced chemical equation: 3CuCl2 * 2H2O + 2Al -> 3Cu + 2AlCl3 + 6H2O.
- From the equation, you can see that the ratio of moles of Cu to moles of CuCl2 * 2H2O is 3:5.
- Calculate the moles of Cu:
Moles of Cu = Moles of CuCl2 * 2H2O * (3 mol Cu / 5 mol CuCl2 * 2H2O) = 0.009302 mol * (3/5) = 0.005591 mol.
4. Calculate the mass of Cu using the molar mass of Cu:
- The molar mass of Cu is 63.55 g/mol.
- Calculate the mass of Cu:
Mass of Cu = Moles of Cu * Molar mass of Cu = 0.005591 mol * 63.55 g/mol = 0.3543 g.
Therefore, the number of grams that should have been produced when 1.25g of copper(II) chloride dihydrate reacts with excess aluminum is approximately 0.3543 grams of Cu.
You're close and your process is right but the math is a little off.
By the way, that's dihydrate and not dehydrate. The formula of the dihydrate is CuCl2.2H2O; therefore, you should have added in 2 mols H2O and not 1. The molar mass of CuCl2.2H2O is closer to 170.
Determine the number of grams that should have been produced when 1.25g of copper (2) chloride dehydrate reacts with excess aluminum
so i'm not really sure how you're supposed to calculate the number of grams with hydrates??
MCuCl2= 63.55 g/mol + 2(35.45 g/mol) = 134.45 g/mol
MH2O= 2(1.01 g/mol) + 16 g/mol = 18.02 g/mol
134.45 + 18.02 = 152.47? (is this even what im supposed to do?)
That should be 134.45 + 2*18.02 = ? approx 170.5
mol CuCl2 *H2O= 1.25g x (1 mol/152.47g)
= 0.008198 mol CuCl2 *H2O
should be CuCl2.2H2O. Then mols CuCl2.2H2O = 1.25/170.5 = ?
mole Cu= 0.008198 mol CuCl2 *H2O x (3 mol Cu/5 mol CuCl2 *H2O)
= 0.004918 mol Cu
This is an error on the factor. It should be mols CuCl2.2H2O x (3 mols Cu/3 mols CuCl2.2H2O) = ?
Then mass Cu = mols Cu x atomic mass Cu = ?
MCu= 63.55 g/mol
mass Cu= 0.004918 mol Cu x (63.55g/1 mol)
= 0.3126 g Cu