calculate the number of g fes04 that will be oxidized by 24ml of a 250N Kmn04 in a solution acidified with sulfuric acid. The unbalanced equation for the reaction.
The answer 0.912g
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To calculate the number of moles of FeSO4 that will be oxidized by the given amount of KMnO4, we first need to balance the equation for the reaction. The unbalanced equation is:
KMnO4 + FeSO4 + H2SO4 → MnSO4 + Fe2(SO4)3 + H2O
Next, we need to determine the stoichiometry of the reaction by comparing the coefficients of KMnO4 and FeSO4. From the balanced equation, we can see that the stoichiometric ratio of KMnO4 to FeSO4 is 1:1.
Now, let's calculate the number of moles of KMnO4:
Molarity (M) = moles/volume (L)
250N = moles/0.024L (since the given volume is 24ml, convert it to liters by dividing by 1000)
moles = 0.024L x 250N = 6 moles
Since the stoichiometric ratio of KMnO4 to FeSO4 is 1:1, there will be an equal number of moles of FeSO4 oxidized. Therefore, 6 moles of FeSO4 will be oxidized by the given amount of KMnO4.
To calculate the mass of FeSO4 that corresponds to 6 moles, we need to find the molar mass of FeSO4:
Fe = 55.845 g/mol
S = 32.06 g/mol
O = 16.00 g/mol (x4)
Total molar mass = 55.845 + 32.06 + (16.00 x 4) = 151.848 g/mol
Formula mass of FeSO4 = 151.848 g/mol
Mass = moles x formula mass
Mass = 6 moles x 151.848 g/mol = 911.088 g
Therefore, the number of grams of FeSO4 that will be oxidized by 24 ml of 250N KMnO4 in a solution acidified with sulfuric acid is approximately 911.088 grams.
To calculate the number of grams of FeSO4 that will be oxidized by the given amount of KMnO4, we need to first balance the equation representing the reaction between FeSO4 and KMnO4, and then apply stoichiometry.
The unbalanced equation for the reaction is:
KMnO4 + FeSO4 + H2SO4 -> MnSO4 + Fe2(SO4)3 + H2O
Now, let's balance the equation by ensuring that the number of atoms on each side of the equation is the same:
2KMnO4 + 10FeSO4 + 8H2SO4 -> 2MnSO4 + 5Fe2(SO4)3 + 8H2O
Next, we need to determine the stoichiometry of the reaction, which relates the ratio of moles between the reactants and the products. From the balanced equation, we can see that:
2 moles of KMnO4 react with 10 moles of FeSO4
Since the concentration of the KMnO4 solution is given as 250N, we can convert it to moles using the formula:
moles of KMnO4 = concentration (N) x volume (L)
Here, the volume of KMnO4 solution is 24 mL, which is equivalent to 0.024 L. Thus, the moles of KMnO4 can be calculated as:
moles of KMnO4 = 250N x 0.024 L = 6 moles
From the stoichiometry, we know that 2 moles of KMnO4 react with 10 moles of FeSO4. Therefore, the moles of FeSO4 required to react with 6 moles of KMnO4 can be calculated as:
moles of FeSO4 = (6 moles KMnO4 x 10 moles FeSO4) / 2 moles KMnO4 = 30 moles
Now, we can calculate the mass of FeSO4 required to react with these 30 moles, using the molar mass of FeSO4:
mass of FeSO4 = moles of FeSO4 x molar mass of FeSO4
The molar mass of FeSO4 can be calculated by adding the atomic masses of the individual elements: Fe (55.845 g/mol) + S (32.06 g/mol) + 4O (16.00 g/mol) = 151.88 g/mol
Therefore, the mass of FeSO4 required is:
mass of FeSO4 = 30 moles x 151.88 g/mol = 4556.4 g
However, the answer you provided is given as 0.912 g. It seems there may be an error in the question or calculation. Please double-check the values provided or share any additional information to help me provide a more accurate answer.
mL x N = # milliequivalents KMnO4.
# m.e. KMnO4 = # m.e. FeSO4
# m.e. FeSO4 x equivalent weight FeSO4 = g FeSO4.
All of that can be put together to make one nice equation.
mL x N x m.e.w. = grams.
24 x 0.250 x 0.152 = 0.0912g FeSO4.