A)How many moles of O2 are required for the complete combustion of 2.2 g of C3H8 to form CO2 and H2O?
b)A 65.25 g sample of CuSO4•5H2O (M = 249.7) is dissolved in enough water to make 0.800 L of solution. What volume of this solution must be diluted with water to make 1.00 L of 0.100 M CuSO4?
306mL
a) To determine the number of moles of O2 required for the complete combustion of C3H8, you need to use the balanced chemical equation for the combustion reaction:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, you can see that 1 mol of C3H8 requires 5 mol of O2 for complete combustion.
First, calculate the molar mass of C3H8:
(3 × 12.01 g/mol) + (8 × 1.01 g/mol) = 44.11 g/mol
Now, calculate the number of moles of C3H8:
moles of C3H8 = mass of C3H8 / molar mass of C3H8
moles of C3H8 = 2.2 g / 44.11 g/mol
moles of C3H8 = 0.0498 mol
According to the balanced equation, 1 mol of C3H8 requires 5 mol of O2. Therefore, the number of moles of O2 required can be calculated as:
moles of O2 = 0.0498 mol C3H8 * (5 mol O2 / 1 mol C3H8)
moles of O2 = 0.249 mol
So, 0.249 moles of O2 are required for the complete combustion of 2.2 g of C3H8.
b) The molar mass of CuSO4•5H2O is 249.7 g/mol.
To calculate the amount of CuSO4•5H2O needed to prepare a 0.100 M solution, first, calculate the number of moles of CuSO4:
moles of CuSO4 = molarity * volume in liters
moles of CuSO4 = 0.100 mol/L * 1.00 L
moles of CuSO4 = 0.100 mol
Now, convert moles of CuSO4 to grams:
mass of CuSO4•5H2O = moles of CuSO4•5H2O * molar mass of CuSO4•5H2O
mass of CuSO4•5H2O = 0.100 mol * 249.7 g/mol
mass of CuSO4•5H2O = 24.97 g
Since the initial sample is 65.25 g, you need to find the volume of this solution that contains 24.970 g of CuSO4•5H2O:
volume of solution containing CuSO4•5H2O = mass of CuSO4•5H2O / molar mass of CuSO4•5H2O
volume of solution containing CuSO4•5H2O = 24.970 g / 249.7 g/mol
volume of solution containing CuSO4•5H2O = 0.0999 mol
To dilute this solution to 1.00 L of 0.100 M CuSO4, you will need to add water to reduce the concentration. Subtract the moles of CuSO4 in the desired final solution from the original volume:
moles of CuSO4 remaining in solution = initial moles - final moles
moles of CuSO4 remaining in solution = 0.0999 mol - 0.100 mol
moles of CuSO4 remaining in solution = -0.0001 mol
Since you cannot have a negative mole value, it means that you need to add water to the solution and dilute it to 1.00 L to achieve the desired concentration of 0.100 M CuSO4.
a) To determine the number of moles of O2 required for the complete combustion of 2.2 g of C3H8, we need to use stoichiometry.
The balanced equation for the combustion of C3H8 is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the balanced equation, we can see that the mole ratio of C3H8 to O2 is 1:5. This means that for every mole of C3H8, we need 5 moles of O2.
First, let's calculate the number of moles of C3H8:
Molar mass of C3H8 = (3 * atomic mass of C) + (8 * atomic mass of H)
= (3 * 12.01 g/mol) + (8 * 1.01 g/mol)
= 36.03 g/mol + 8.08 g/mol
= 44.11 g/mol
Number of moles of C3H8 = mass of C3H8 / molar mass of C3H8
= 2.2 g / 44.11 g/mol
≈ 0.0499 mol
Now, we can determine the number of moles of O2 required:
Number of moles of O2 = number of moles of C3H8 * mole ratio (O2/C3H8)
= 0.0499 mol * 5
= 0.2495 mol
Therefore, approximately 0.2495 moles of O2 are required for the complete combustion of 2.2 g of C3H8.
b) To determine the volume of the CuSO4•5H2O solution that must be diluted to make 1.00 L of 0.100 M CuSO4, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molarity of solution 1
V1 = initial volume of solution 1
M2 = final molarity of solution 2
V2 = final volume of solution 2
In this case, we are given:
M1 = molarity of CuSO4•5H2O solution = ?
V1 = initial volume of CuSO4•5H2O solution = 65.25 g / molar mass of CuSO4•5H2O (249.7 g/mol) = 0.2616 mol
M2 = final molarity of 0.100 M CuSO4 solution = 0.100 M
V2 = final volume of 0.100 M CuSO4 solution = 1.00 L
Rearranging the equation, we can solve for M1:
M1 = (M2 * V2) / V1
= (0.100 M * 1.00 L) / (0.2616 mol)
≈ 0.3822 M
Now that we have the initial molarity, we can find the initial volume of the solution that needs to be diluted.
V1 = (M2 * V2) / M1
= (0.100 M * 1.00 L) / (0.3822 M)
≈ 0.2618 L
Therefore, approximately 0.2618 L of the CuSO4•5H2O solution must be diluted with water to make 1.00 L of 0.100 M CuSO4 solution.
A)
C3H8 + 5O2 ==> 3CO2 + 4H2O
mols C3H8 = 2.2 g/molar mass C3H8.
Convert mols C3H8 to mols CO2.
b)
mols CuSO4.5H2O = g/molar mass
M = mols/L soln.
Then use the dilution formula of
c1v1 = c2v2
c = concn
v = volume