Calculate the amount of 6 M nitric acid you need to add to fully react with .5 grams of copper and add a 20 percent excess.
Here is a worked example of a stoichiometry problem such as this. After finding moles HNO3, the M = moles/L will give you volume HNO3, then add 20% excess to that.
http://www.jiskha.com/science/chemistry/stoichiometry.html
I still do not understand.
You haven't had time to try yet. Follow the steps one by one. Those instructions, which I wrote, will take you to the answer.
The equation is
3Cu + 8HNO3 ==>3Cu(NO3)2 + 2NO + 4H2O
Check that to make sure it is balanced.
It still doesnt make aense because the answer comes out to 240 liters.
That's not a balanced equation. It should be Cu + 4HNO3 ==>Cu(NO3)2 + 2NO + 2H2O
To calculate the amount of 6 M nitric acid needed to fully react with 0.5 grams of copper and provide a 20% excess, we need to follow a few steps:
Step 1: Determine the molar mass of copper (Cu).
Copper (Cu) has a molar mass of approximately 63.55 g/mol.
Step 2: Convert the mass of copper to moles.
Using the formula: moles = mass / molar mass
moles = 0.5 g / 63.55 g/mol
moles ≈ 0.00787 mol (rounded to five decimal places)
Step 3: Calculate the balanced chemical equation.
The balanced chemical equation for the reaction between copper and nitric acid (HNO3) is:
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
This equation tells us that for every 3 moles of copper, 8 moles of nitric acid are required for a complete reaction.
Step 4: Calculate the moles of nitric acid required for the amount of copper given.
moles of nitric acid required = moles of copper x (8 moles HNO3 / 3 moles Cu)
moles of nitric acid required = 0.00787 mol x (8 / 3)
moles of nitric acid required ≈ 0.020966667 mol (rounded to nine decimal places)
Step 5: Determine the volume of 6 M nitric acid needed.
The concentration of nitric acid is given as 6 M (meaning 6 moles per liter).
volume of nitric acid in liters = moles of nitric acid required / concentration
volume of nitric acid in liters = 0.020966667 mol / 6 mol/L
volume of nitric acid in liters ≈ 0.003494444 L (rounded to nine decimal places)
Step 6: Convert the volume of nitric acid to milliliters (ml).
volume of nitric acid in ml = volume of nitric acid in liters x 1000
volume of nitric acid in ml ≈ 0.003494444 L x 1000
volume of nitric acid in ml ≈ 3.494444 ml (rounded to six decimal places)
Step 7: Add the excess.
To add a 20% excess, multiply the calculated volume of nitric acid by 1.2.
excess volume of nitric acid in ml = volume of nitric acid in ml x 1.2
excess volume of nitric acid in ml ≈ 3.494444 ml x 1.2
excess volume of nitric acid in ml ≈ 4.193333 ml (rounded to six decimal places)
Therefore, you would need approximately 4.193333 ml of 6 M nitric acid to fully react with 0.5 grams of copper while providing a 20% excess.