A 5.0 g sample of Cu(NO3)2-nH20 is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?
I got as far as finding the molar mass of Cu(NO3)2, which is 187.5g/mol
Please help me further with this question.
5 answers
Also please explain every step.
g Cu(NO3)2 = 3.9
g H2O = 5.0-3.9 = 1.1
mols Cu(NO3)2 = 3.9/187.5 = 0.0208
mols H2O = 1.1/18 = 0.0611
You want n mols H2O/1 mol Cu(NO3)2
So 1 mol Cu(NO3)2 = 0.0208/0.0208 = 1.000
n mols H2O = 0.0611/0.0208 = 2.94. Round that to a whole number of 3.0
Formula is Cu(NO3)2*3H2O
g H2O = 5.0-3.9 = 1.1
mols Cu(NO3)2 = 3.9/187.5 = 0.0208
mols H2O = 1.1/18 = 0.0611
You want n mols H2O/1 mol Cu(NO3)2
So 1 mol Cu(NO3)2 = 0.0208/0.0208 = 1.000
n mols H2O = 0.0611/0.0208 = 2.94. Round that to a whole number of 3.0
Formula is Cu(NO3)2*3H2O
the answer is 2.9 converted into 3
I didn't understand....please explain in detail
I also did not understand.Please make it in a little bit easier way:)