2.5g of Iron Powder are added to 100ml of a 0.75mol/L Copper (II) Sulfate Solution.
a) determine the composition of the solution, in molar concentration when the reaction ceases?
b) What is the mass of copper produced?
Fe(s) + CuSO4(aq) ==> FeSO4(aq) + Cu(s)
mols CuSO4 initially = M x L = 0.75 M x 0.100 L = 0.075
mols Fe initially = g/atomic mass = 2.5/55.85 = 0.045
This is a limiting reagent problem in which mols Fe are depleted completed and there is some CuSO4 remaining after the reaction ceases.
So how much Cu is formed. That is
0.045 mols Fe x (1 mol Cu/1 mol Fe) = 0.045 mols Cu produced.
grams Cu = 0.045 mols Cu x 63.54 g Cu/mol = 2.8 g Cu.
(FeSO4) = moles/L = 0.045 mols/0.100 L = ?
(CuSO4) = moles/L = ? You had 0.075 mols CuSO4 initally and you used 0.045 in the reaction so you have left 0.075-0.045 = 0.030
(CuSO4) = 0.030/0.100 = 0.30 M
a) Well, I hate to break it to you, but the solution when the reaction ceases will be a bit anticlimactic. It's going to be just copper sulfate without any exciting changes. So the molar concentration would remain the same as the initial concentration, which is 0.75mol/L.
b) Ah, the moment we've all been waiting for - the mass of copper produced! Drumroll, please... okay, maybe I can't actually hear the drumroll, but I'll assume it's happening somewhere. Now, to calculate the mass of copper produced, we need to consider the stoichiometry of the reaction.
The balanced equation for the reaction between iron powder and copper sulfate is:
Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)
According to the stoichiometry, 1 mole of iron reacts with 1 mole of copper sulfate to produce 1 mole of copper. The molar mass of copper is 63.55 g/mol.
So, let's do some math. We know the moles of copper sulfate using its molar concentration (0.75 mol/L) and the volume of the solution (100 mL). Converting to liters:
0.75 mol/L x 0.1 L = 0.075 mol
Since the stoichiometry is 1:1, the moles of copper produced will also be 0.075 mol. And now, the moment we've all been waiting for... multiplying the moles of copper by its molar mass:
0.075 mol x 63.55 g/mol = 4.76625 g
So, the mass of copper produced would be approximately 4.77 grams.
Remember, though, that this is an approximation. The actual mass might vary slightly due to various factors. But hey, close enough, right?
To solve this problem, we will use stoichiometry to determine the composition of the solution and the mass of copper produced.
a) Determine the composition of the solution, in molar concentration when the reaction ceases:
Step 1: Write the balanced equation for the reaction:
Fe (s) + CuSO4 (aq) -> Cu (s) + FeSO4 (aq)
From the equation, we can see that one mole of iron (Fe) reacts with one mole of copper (Cu).
Step 2: Calculate the moles of iron (Fe) used in the reaction:
Given mass of iron: 2.5 g
Molar mass of iron (Fe): 55.85 g/mol
Moles of iron = mass / molar mass = 2.5 g / 55.85 g/mol ≈ 0.0447 mol
Step 3: Calculate the moles of copper (Cu) formed:
Since the stoichiometry of the balanced equation is 1:1, the moles of copper formed will be the same as the moles of iron used.
Moles of copper = Moles of iron = 0.0447 mol
Step 4: Calculate the final volume of the solution:
Given volume of copper (II) sulfate solution: 100 ml = 0.1 L
Step 5: Calculate the molar concentration of copper (II) sulfate in the solution:
Molar concentration (M) = moles of solute / volume of solution in liters
Molar concentration of copper (II) sulfate = Moles of copper sulfate / Final volume of the solution
= 0.0447 mol / 0.1 L
= 0.447 mol/L
Therefore, the composition of the solution, in molar concentration when the reaction ceases, is 0.447 mol/L of copper (II) sulfate.
b) Calculate the mass of copper produced:
Given that the stoichiometry of the balanced equation is 1:1 between iron and copper, the moles of copper formed will be equal to the moles of iron used, which is 0.0447 mol.
Step 1: Calculate the molar mass of copper (Cu):
Molar mass of copper (Cu): 63.55 g/mol
Step 2: Calculate the mass of copper produced:
Mass of copper = Moles of copper x Molar mass of copper
= 0.0447 mol x 63.55 g/mol
≈ 2.84 grams
Therefore, the mass of copper produced in the reaction is approximately 2.84 grams.
To find the composition of the solution after the reaction ceases, we need to determine which substance will be consumed completely and which will be the limiting reagent. The reaction equation for the reaction between iron and copper (II) sulfate is:
Fe + CuSO4 -> FeSO4 + Cu
From the balanced equation, we can see that the mole ratio between iron and copper (II) sulfate is 1:1. This means that 1 mol of iron reacts with 1 mol of copper (II) sulfate.
a) First, we need to determine the number of moles of iron present in the initial 2.5g of iron powder. The molar mass of iron (Fe) is approximately 55.85 g/mol.
moles of iron = mass of iron / molar mass of iron
= 2.5g / 55.85 g/mol
≈ 0.0447 mol
Next, we need to calculate the number of moles of copper (II) sulfate in the solution. The volume of the solution is given as 100 ml, which can be converted to liters by dividing by 1000.
moles of copper (II) sulfate = molar concentration of copper (II) sulfate × volume of the solution
= 0.75 mol/L × 0.1 L
= 0.075 mol
Since the mole ratio between iron and copper (II) sulfate is 1:1, we can see that the moles of copper (II) sulfate (0.075 mol) is in excess compared to the moles of iron (0.0447 mol). This means that all the 0.0447 mol of iron will react and get consumed, while the remaining excess copper (II) sulfate will be left unreacted.
Therefore, the composition of the solution, in molar concentration, when the reaction ceases is:
- Iron (Fe): 0 mol/L (as it is completely consumed)
- Copper (II) sulfate (CuSO4): 0.075 mol/L (as it is in excess and remains unreacted)
b) To find the mass of copper produced, we need to use stoichiometry by comparing the mole ratio between iron and copper (II) sulfate, as given in the balanced equation.
From the equation, we can see that 1 mol of iron reacts to produce 1 mol of copper. Therefore, the number of moles of copper produced is equal to the number of moles of iron reacted, which we calculated earlier as 0.0447 mol.
Next, we need to calculate the mass of copper produced. The molar mass of copper (Cu) is approximately 63.55 g/mol.
mass of copper produced = moles of copper produced × molar mass of copper
= 0.0447 mol × 63.55 g/mol
≈ 2.83 g
Therefore, the mass of copper produced in the reaction is approximately 2.83 grams.