Suppose that you used 0.75g of iron. What is the minimum volume of 1.5 M_ CuSO4 solution required
Using eqn. (3-1)? and eqn. (3-2)?
(3-1)
Fe(s) + Cu^2+ (aq) ---- Fe^2 (aq)+ Cu (s)
(3-2)
2Fe(s)+ 3Cu^2+(aq) --- 2Fe^3+ (aq) +3Cu (s)
You don't provide enough information for me to know if you have both equations or just one reacting.
If the first one only, it is
(0.75/55.85) x (1 mol Cu^2+/1 mol Fe) = ?; then M CuSO4 = mols CuSO4/L CuSO4. You know mols and M, solve for L.
If the second one only, it is
(0.75/55.85) x (3 mol Cu^2+/2 mol Fe) =?; then, M CuSO4 = mols CuSO4/L CuSO4. You know mols and M, solve for L CuSO4.
If both then add the volumes CuSO4 together.
Also, if you can provide more information concerning the lab experiment, I may be able to distinguish betwen the two options.
To find the minimum volume of the 1.5 M CuSO4 solution required, we need to use equation (3-1) and equation (3-2) and consider the stoichiometry of the reactions.
First, let's calculate the number of moles of iron (Fe) used. We have 0.75 g of iron.
The molar mass of iron (Fe) is approximately 55.85 g/mol.
Number of moles of Fe = Mass / Molar mass
Number of moles of Fe = 0.75 g / 55.85 g/mol ≈ 0.01342 mol
Now, let's examine equation (3-1) and equation (3-2) to determine the stoichiometry of CuSO4.
From equation (3-1), we can see that the stoichiometric ratio between Fe and Cu^2+ is 1:1.
From equation (3-2), we can see that the stoichiometric ratio between Fe and Cu^2+ is 2:3.
Comparing these two ratios, we can see that equation (3-2) requires twice as much Fe as equation (3-1) to react with the same amount of Cu^2+.
Therefore, equation (3-2) will be the limiting reaction because it requires more Fe.
To balance the stoichiometry, we need to determine how many moles of Cu^2+ are required for equation (3-2) using the ratio from the equation.
From equation (3-2), the stoichiometric ratio between Fe and Cu^2+ is 2:3.
Number of moles of Cu^2+ = (2/3) * 0.01342 mol ≈ 0.00895 mol
Finally, we can calculate the minimum volume of the 1.5 M CuSO4 solution required.
To do this, we'll use the equation:
Volume (in L) = Number of moles / Molarity
For equation (3-2), we need to calculate the volume of the 1.5 M CuSO4 solution required.
Volume = 0.00895 mol / 1.5 mol/L ≈ 0.00597 L or 5.97 mL
Therefore, the minimum volume of the 1.5 M CuSO4 solution required is approximately 5.97 mL.
To find the minimum volume of 1.5 M CuSO4 solution required, we need to determine the limiting reactant between Fe(s) and Cu^2+ (aq) based on the given amount of iron used (0.75g).
Let's start with equation (3-1):
Fe(s) + Cu^2+ (aq) ---- Fe^2 (aq) + Cu (s)
From the balanced equation, we see that 1 mole of Fe reacts with 1 mole of Cu^2+ to produce 1 mole of Fe^2+ and 1 mole of Cu. The molar mass of iron (Fe) is approximately 56 g/mol.
To find the number of moles of Fe, we use the formula:
moles = mass / molar mass
moles of Fe = 0.75 g / 56 g/mol
moles of Fe = 0.01339 mol
Since the stoichiometry of the reaction is 1:1 between Fe and Cu^2+, we can conclude that we need 0.01339 moles of Cu^2+ to fully react with the given amount of Fe.
Now let's move on to equation (3-2):
2Fe(s) + 3Cu^2+(aq) --- 2Fe^3+ (aq) + 3Cu (s)
From the balanced equation, we see that 2 moles of Fe react with 3 moles of Cu^2+ to produce 2 moles of Fe^3+ and 3 moles of Cu.
Since we determined that we need 0.01339 moles of Cu^2+ from equation (3-1), we divide this value by the stoichiometric ratio of Cu^2+ to Cu (3:3) to find the number of moles of Cu needed:
moles of Cu = 0.01339 mol / 3
moles of Cu = 0.004463 mol
Now that we know the moles of Cu needed, we can calculate the volume of the 1.5 M CuSO4 solution required using the equation:
moles = concentration * volume
volume = moles / concentration
volume = 0.004463 mol / 1.5 M
volume = 0.002975 L or 2.975 mL
Therefore, the minimum volume of 1.5 M CuSO4 solution required, using eqn. (3-1) and eqn. (3-2), is approximately 2.975 mL.