[Lindsay]

Hello Guys, I'm taking Gen Chemistry. I really like it a lot. Except that in my Lab Book there is a problem that I do not know how to solve.
Here it is,
What volume, in drops, of 16 M (conc) HNO3 is required to react with 0.0214 g of Cu metal. Assume 20 drops per milliliter.
That is it!
I went ahead and did the following;

Cu +4 HNO3 yields Cu(NO3)2 + 2NO2 + 2H2O

molar mass for HNO3 = 63g/mol
mass of HNO3 in 1 L solution : 16 mole X 63 g/mol =1008 g

0.0214g Cu X 1 mole CU/63.55 g Cu X 4 mole HNO3/1 mole Cu...

I don't think that is right, please help.

Lindsay

[enahs]

0.0214g Cu X 1 mole CU/63.55 g Cu X 4 mole HNO3/1 mole Cu...

I don't think that is right, please help.

But that part I quoted is correct! So it tells you how many mols of HNO₃ you need, correct?
So what volume of HNO3 gives you that many mols?
How many drops is that volume?

[Lindsay]

Thanks for your response,

Will this be the answer;

0.0214g Cu X 1 mol CU/63.55gCu X 4 mol HNO3/1 mol Cu
= 0.00135 mol HNO3 X 1008 g HNO3 = 1.36g HNO3 X 1L/1000g X 1000ML/1L
Now, I get milliliter = 1.36ML X 20 Drops/1ML = 27.2 Drops

Please let me know

Thanks,
Lindsay

[Borek]

Number of moles of acid is correct, but then you are doing some strange things. You don't need to convert it to mass, you just need to find out volume in mL.

Besides, no idea what 1008 g is.

[Lindsay]

What volume, in drops, of 16 M (conc) HNO3 is required to react with 0.0214 g of Cu metal. Assume 20 drops per milliliter.
That is it!
The problem is;

I went ahead and did the following;

Cu +4 HNO3 yields Cu(NO3)2 + 2NO2 + 2H2O

molar mass for HNO3 = 63.02g/mol
mass of HNO3 in 1 L solution : 16 mole X 63.02 g/mol =1008 g ( This is where I get 1008g of HNO3)

My result is

0.0214g Cu X 1 mol CU/63.55gCu X 4 mol HNO3/1 mol Cu
= 0.00135 mol HNO3 X 1008 g HNO3 = 1.36g HNO3 X 1L/1000g X 1000ML/1L
Now, I get milliliter = 1.36ML X 20 Drops/1ML = 27.2 Drops

If I don't use the 1008, then
0.0214g Cu X 1 mol CU/63.55gCu X 4 mol HNO3/1 mol Cu
= 0.00135 mol HNO3 X 63.02g HNO3/ 1 mol HNO3 X 1L/ 1000g X 1000ML/1L
=0.849 ML HNO3 X 20 Drops = 1.70 Drops

Is the answer 1.70 Drops of HNO3

Thanks,

Lindsay

[Lindsay]

Please disregard the previous post. I think I figured it out,


My result is

0.0214g Cu X 1 mol CU/63.55gCu X 4 mol HNO3/1 mol Cu
= 0.00135 mol HNO3 X 1 L HNO3 / 16 moles HNO3) = 8.42 X 10 (-5) L X 1000ML/1L
= 0.0841 ML  X 20Drops/1ML = 1.68 Drops 16 M HNO3

Did I do it right this time?
Please advise,

Lindsay

[Borek]

I have already prepared long and boring answer trying to explain what and why you did wrong, but before I have hit Post you did it OK

[Lindsay]

Cool!!!!!!! I appreciate your help.