using your initial mass of copper, calculate the minimum volume of concentrated 12M nitric acid needed to completely react with your copper to form copper(II) nitrate? Original mass of copper was 0.40g.
I will assume you are using concentrated HNO3. With concd HNO3 the reduction product is NO2; with dilute HNO3 the reduced product is NO.
Cu + 4HNO3 ==> Cu(NO3)2 + 2H2O + 2NO2
mols Cu = grams/molar mass = approx 0.0063 but you need to do it and the calculations that follow more accurately because I've estimated the numbers.
Convert mols Cu to mols HNO3. That's 0.0063 x (4 mol HNO3/1 mol Cu) = approx 0.025
Then M HNO3 = mols HNO3/L HNO3. You know M HNO3 from the problem and mols from the above calculation; solve for L: HNO3 and convert to mL.
Ah, chemistry calculations! Brace yourself for a clownish adventure. To find the minimum volume of 12M nitric acid needed to react with your 0.40g of copper, let's embark on this humorous journey.
First, we need to convert the mass of copper to moles. Copper has a molar mass of approximately 63.55 g/mol. So, using our math skills, we find that 0.40g of copper is around 0.00628 moles. Lovely!
Next, let's utilize the balanced equation for the reaction between copper and nitric acid:
3Cu + 8HNO3 -> 3Cu(NO3)2 + 4H2O + 2NO
From a glimpse at this equation, we can see that our ratio is 3:8. We need 3 moles of copper to react with 8 moles of nitric acid. Since we have 0.00628 moles of copper, we'll need (0.00628 moles * 8 moles of nitric acid) / 3 moles of copper, which comes out to approximately 0.01674 moles of nitric acid.
To determine the minimum volume of 12M nitric acid needed, we'll divide 0.01674 moles by the acid's Molarity (12M). This gives us 0.001395 L or 1.4 mL.
So, there you have it (in a not-so-serious way), the minimum volume of concentrated 12M nitric acid needed to react with your 0.40g of copper is roughly 1.4 mL. Have a blast with your chemistry experiments!
To calculate the minimum volume of concentrated 12M nitric acid needed to react with 0.40g of copper, we can follow these steps:
Step 1: Convert the mass of copper to moles.
To do this, we need the molar mass of copper. The atomic mass of copper is 63.55 g/mol. Therefore, the number of moles of copper is calculated as follows:
moles of copper = mass of copper / molar mass of copper
= 0.40g / 63.55 g/mol
≈ 0.00629 moles of copper
Step 2: Determine the stoichiometry of the balanced chemical equation.
From the balanced chemical equation, we can see that 3 moles of copper react with 8 moles of nitric acid to form 3 moles of copper(II) nitrate.
Step 3: Calculate the moles of nitric acid required.
From the stoichiometry of the balanced chemical equation, we know that the ratio of moles of copper to moles of nitric acid is 3:8. Therefore, the moles of nitric acid required can be calculated as follows:
moles of nitric acid = (moles of copper) x (moles of nitric acid / moles of copper)
= 0.00629 moles × (8 moles / 3 moles)
≈ 0.0168 moles of nitric acid
Step 4: Calculate the volume of nitric acid required.
We know the concentration of the nitric acid is 12M (12 moles per liter).
volume of nitric acid = (moles of nitric acid) / (concentration of nitric acid)
= 0.0168 moles / 12 M
≈ 0.00140 L or 1.40 mL
Therefore, the minimum volume of concentrated 12M nitric acid needed to completely react with 0.40g of copper to form copper(II) nitrate is approximately 1.40 mL.
To determine the minimum volume of concentrated 12M nitric acid required to react completely with 0.40g of copper, we need to follow several steps:
Step 1: Convert the mass of copper to moles.
To convert the mass of copper to moles, we need to use the molar mass of copper. The molar mass of copper (Cu) is 63.55 g/mol.
Number of moles of copper = mass / molar mass
= 0.40g / 63.55 g/mol
≈ 0.0063 mol
Step 2: Determine the stoichiometry of the chemical reaction.
From the given information, the reaction is as follows:
Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
According to the balanced equation, 1 mole of copper reacts with 4 moles of nitric acid (HNO3) to produce 1 mole of copper(II) nitrate (Cu(NO3)2).
Step 3: Calculate the moles of nitric acid needed.
The molar ratio between copper and nitric acid is 1:4. Therefore, the moles of nitric acid needed can be calculated as follows:
Moles of nitric acid required = moles of copper × (4 moles of nitric acid/1 mole of copper)
= 0.0063 mol × (4/1)
= 0.0252 mol
Step 4: Convert moles of nitric acid to volume.
To convert the moles of nitric acid to volume, we need to use its molarity. The concentration of nitric acid is given as 12M, which means 12 moles of nitric acid are present in 1 liter (1000 mL) of solution.
Volume of nitric acid required = (moles of nitric acid required) / (concentration of nitric acid in moles per liter)
= 0.0252 mol / 12 mol/L
≈ 0.0021 L or 2.1 mL
Therefore, the minimum volume of concentrated 12M nitric acid needed to completely react with 0.40g of copper to form copper(II) nitrate is approximately 2.1 mL.