Students are instructed to add no more than 0.25 mL of 15 M nitric acid. What
volume of 15 M nitric acid is required to react with 0.030 g of copper metal? If 1.0 mL of acid
contains approximately 20 drops, how many drops of nitric acid are needed?
you have to first calculate the moles of the copper, then 0.03/63.01 then afterwards times it by 0.25 because thats how much ml you need, then times it by 20 to find the drops.
Do you know the erquation. 15 M HNO2 is strong, look on Google to find the reaction with concentrated HNO3.
Wow, yall were no help. I just needed help and none of these replies look remotely close to what I got.
To find the volume of 15 M nitric acid required to react with 0.030 g of copper metal, we need to use stoichiometry.
1. Start by writing the balanced chemical equation for the reaction between nitric acid (HNO3) and copper (Cu):
8HNO3 + 3Cu -> 3Cu(NO3)2 + 2NO + 4H2O
2. Determine the molar mass of copper (Cu). The atomic mass of copper is 63.55 g/mol.
3. Convert the mass of copper to moles using the formula:
moles = mass / molar mass
moles = 0.030 g / 63.55 g/mol
moles ≈ 0.000472 mol
4. From the balanced equation, we can see that 3 moles of Cu react with 8 moles of HNO3. Therefore, the ratio is:
3 mol Cu / 8 mol HNO3
5. Calculate the volume of 15 M nitric acid needed using the formula:
volume (in L) = moles of HNO3 / concentration of HNO3
volume (in L) = (0.000472 mol / 15 mol/L)
volume (in L) ≈ 0.000031 L
6. Finally, convert the volume to milliliters (mL):
volume (in mL) = volume (in L) * 1000
volume (in mL) ≈ 0.031 mL
To find the number of drops of nitric acid needed, we need to know how many drops are in 1 mL of acid, which is given as 20 drops.
Therefore, the number of drops of nitric acid needed is:
drops = volume (in mL) * 20
drops ≈ 0.031 mL * 20
drops ≈ 0.62 drops
So, approximately 0.62 drops of nitric acid are needed.