Calculate the volume of 1.0 M copper(II) sulfate solution required to completely react with 1.0 g of Fe(s) according to this equation: 2 Fe(s) + 3 Cu^2+ (aq) --> 2 Fe^3+ (aq) + 3 Cu (s).
the copper in the reactant has a charge of 2+ and the iron in the product has a charge of 3+..sorry for the confusion. can anyone walk me through this? I'm unsure where to begin.
mols Fe = grams/atomic mass
Use the coefficients in the balanced equation to convert mols Fe to mols Cu^2+ (i.e. CuSO4).
Then M CuSO4 = mols CuSO4/L CuSO4. You have M and mols, solve for L. Convert to mL if needed.
10 mL
1.0
Well, well, well, looks like we've got a chemistry query here! Fear not, my friend, Clown Bot is here to help you out with some humorous guidance.
Firstly, let's start by figuring out the molar mass of copper(II) sulfate. Copper (Cu) has an atomic mass of around 63.5 g/mol, sulfur (S) has a mass of approximately 32.1 g/mol, and four oxygen (O) atoms have a combined mass of roughly 64.0 g/mol. So, the molar mass of copper(II) sulfate is around 159.5 g/mol.
Now, with that fact in mind, let's calculate the number of moles of Fe(s) you have. As we know, the molar mass of iron (Fe) is approximately 55.8 g/mol. So, 1.0 g of Fe(s) is equal to 1.0 g / 55.8 g/mol ≈ 0.018 moles of Fe(s).
According to the balanced equation, we require 3 moles of Cu^2+ for every 2 moles of Fe(s) to completely react. So, for 0.018 moles of Fe(s), we'll need (3/2) × 0.018 ≈ 0.027 moles of Cu^2+.
Now, let's determine the volume of the 1.0 M Cu^2+ solution needed to contain 0.027 moles of Cu^2+. Since the concentration is given in Molarity (M), we use the equation:
Molarity (M) = moles / volume (in liters).
Rearranging the equation, we find:
volume (in liters) = moles / Molarity (M).
So, the volume of the 1.0 M Cu^2+ solution required is 0.027 moles / 1.0 M ≈ 0.027 liters, which is equal to 27 milliliters (mL).
Therefore, you would need approximately 27 mL of the 1.0 M copper(II) sulfate solution to completely react with 1.0 g of Fe(s) according to the given equation.
I hope this helps, and remember, even chemistry problems can be fun when you have a clown bot by your side!
To calculate the volume of 1.0 M copper(II) sulfate solution required to completely react with 1.0 g of Fe(s), we can follow these steps:
Step 1: Determine the molar mass of Fe(s)
We need to find the molar mass of Fe(s), which is the atomic mass of iron (Fe) from the periodic table. The atomic mass of Fe is approximately 56 g/mol.
Step 2: Calculate the moles of Fe(s)
To calculate the moles of Fe(s), we use the formula:
moles = mass / molar mass
In this case, the mass of Fe(s) is given as 1.0 g. So the moles of Fe(s) = 1.0 g / 56 g/mol.
Step 3: Use the stoichiometry of the balanced equation
Looking at the balanced equation: 2 Fe(s) + 3 Cu^2+ (aq) -> 2 Fe^3+ (aq) + 3 Cu (s), we can see that 2 moles of Fe(s) react with 3 moles of Cu^2+ (aq).
Step 4: Calculate the moles of Cu^2+ (aq)
Since there is a 2:3 ratio between Fe(s) and Cu^2+ (aq), we can calculate the moles of Cu^2+ (aq) using the relation:
moles of Cu^2+ (aq) = (moles of Fe(s) / 2) * (3 moles of Cu^2+ / 2 moles of Fe(s))
Step 5: Determine the volume of 1.0 M CuSO4 solution
Now that we know the moles of Cu^2+ (aq), we can calculate the volume of the 1.0 M CuSO4 solution using the relation:
volume (in liters) = moles / molarity
In this case, the moles of Cu^2+ (aq) will be equal to the moles of Fe(s). So the volume of the 1.0 M CuSO4 solution = (moles of Fe(s) / 2) * (3 moles of Cu^2+ / 2 moles of Fe(s)) / 1.0 M.
Once you have all the values plugged into the equation, calculate the answer to find the volume of the 1.0 M copper(II) sulfate solution required.