What volume, in drops, of concentrated HNO3 (6M) is required to react with 0.0191 g of Cu metal (20 drops per milliliter)?
The first thing that you need to do is write out the reaction:
Cu + 4HNO3 --> Cu(NO3)2 + 2NO2 + 2H20
Next you want to convert the mass of copper to moles:
Molar mass of copper = 63.54 g/mol
Moles = Mass/(Molar Mass) = 0.0191g / 63.54g/mol = 0.0003 moles
Next you want to multiply by 4 to determine the number of moles of HN03 needed: 0.0012 moles
As stated in the question the solution is 6 moles/L or 0.006 moles / mL. Divide that by the 20drops/mL and you have a concentration of 0.0003 moles/drop
If we divide the number of moles needed, by the number of moles per drop we can find out how many drops of HN03 are needed: 0.0012 / 0.0003 = 4
Therefore, you need 4 drops of HNO3
The reaction of Cu metal with HNO3 is
3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O (This is the reaction of Cu with dilute HNO3, 6M HNO3 is dilute acid)
This means that 8 moles of HNO3 are required for every 3 moles of Cu.
Cu has an atomic weight of 63.546.
0.0191 g of Cu is 0.0191/63.456 or very close to 0.0003 moles. This means we need 0.0008 moles of HNO3.
As the concentration of HNO3 is 6M, one liter of HNO3 has 6 moles. Or 1mL has .006 moles. As 20 drops make a mL 1 drop has .006/20 moles.
Here we need 0.0008 moles of HNO3. Therefore we need .006/(20*.0008)=2.666 drops of HNO3.