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
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
2 answers
Wow, thank you so much that was very descriptive, I really appreciate it!
1 answer