How many milliliters of 0.250 M KMnO4 are needed to react with 3.36 g of Iron(II) sulfate, FeSO4? The reaction is as follows:
10FeSO4(aq) + 2KMnO4(aq) + 8H2SO4(aq) �¨ 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8 H2O(l
To determine the volume of 0.250 M KMnO4 needed to react with 3.36 g of FeSO4, we need to use stoichiometry, the molar mass, and the balanced chemical equation provided.
1. Determine the molar mass of FeSO4:
FeSO4 = (1 x atomic mass of Fe) + (1 x atomic mass of S) + (4 x atomic mass of O)
FeSO4 = (1 x 55.845 g/mol) + (1 x 32.06 g/mol) + (4 x 16.00 g/mol)
FeSO4 = 55.845 g/mol + 32.06 g/mol + 64 g/mol
FeSO4 = 151.905 g/mol
2. Convert the mass of FeSO4 to moles:
Moles of FeSO4 = Mass of FeSO4 / Molar mass of FeSO4
Moles of FeSO4 = 3.36 g / 151.905 g/mol
Moles of FeSO4 ≈ 0.0221 mol
3. Use the stoichiometry of the balanced chemical equation to determine the mole ratio between FeSO4 and KMnO4:
According to the balanced equation, 10 moles of FeSO4 react with 2 moles of KMnO4.
This means that the mole ratio of FeSO4:KMnO4 is 10:2 or 5:1.
4. Determine the moles of KMnO4 needed:
Moles of KMnO4 = Moles of FeSO4 x (1 mole of KMnO4 / 5 moles of FeSO4)
Moles of KMnO4 = 0.0221 mol x (1 mol / 5 mol)
Moles of KMnO4 = 0.00442 mol
5. Calculate the volume of 0.250 M KMnO4 using the following equation:
Volume (L) = Moles / Concentration (M)
Volume (L) = 0.00442 mol / 0.250 mol/L
Volume (L) ≈ 0.0177 L or 17.7 mL
Therefore, approximately 17.7 milliliters of 0.250 M KMnO4 are needed to react with 3.36 g of FeSO4.
To find out how many milliliters of 0.250 M KMnO4 are needed to react with 3.36 g of FeSO4, we need to use stoichiometry and convert the mass of FeSO4 to moles, and then use the balanced chemical equation to determine the ratio between FeSO4 and KMnO4.
Here's the step-by-step process to solve the problem:
1. Calculate the molar mass of FeSO4:
Fe + S + (4 × O) = 55.85 g/mol + 32.06 g/mol + (4 × 16.00 g/mol) = 151.91 g/mol
2. Convert the mass of FeSO4 to moles:
Moles = mass / molar mass = 3.36 g / 151.91 g/mol = 0.0221 mol
3. Use the balanced chemical equation to determine the mole ratio between FeSO4 and KMnO4:
10FeSO4 : 2KMnO4
This means that for every 10 moles of FeSO4, we need 2 moles of KMnO4. Therefore, the ratio between FeSO4 and KMnO4 is 10:2 or 5:1.
4. Determine the number of moles of KMnO4 needed:
Moles of KMnO4 = Moles of FeSO4 / Ratio = 0.0221 mol / (5/1) = 0.00442 mol
5. Finally, use the molarity of KMnO4 to calculate the volume in milliliters:
Molarity (M) = Moles / Volume (L)
Rearranging the equation, we have Volume (L) = Moles / Molarity
Volume (mL) = Volume (L) × 1000 (to convert L to mL)
Volume (mL) = (0.00442 mol / 0.250 mol/L) × 1000 = 17.68 mL
Therefore, 17.68 milliliters of 0.250 M KMnO4 are needed to react with 3.36 g of FeSO4.
Follow the steps in this example. Post your work if you get stuck.
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