I need help setting up this problem.
How many milliliters of 0.203 M KMnO4 are needed to react with 3.64 g of iron(II) sulfate, FeSO4? The reaction is as follows.
10 FeSO4(aq) + 2 KMnO4(aq) + 8 H2SO4(aq) → 5 Fe2(SO4)3(aq) + 2 MnSO4(aq) + K2SO4(aq) + 8 H2O(l)
Thanks
To solve this problem, you will need to use stoichiometry. Stoichiometry is a mathematical tool used in chemistry to relate the amounts of substances involved in a chemical reaction.
Here are the step-by-step instructions to set up this problem:
Step 1: Determine the molar mass of FeSO4.
The molar mass of FeSO4 is calculated by adding up the atomic masses of each element in the compound.
FeSO4 = (1 x atomic mass of Fe) + (1 x atomic mass of S) + (4 x atomic mass of O)
Step 2: Convert the mass of FeSO4 to moles.
To do this, divide the mass of FeSO4 (given as 3.64 g) by its molar mass.
Step 3: Use the balanced chemical equation to determine the stoichiometric ratio between FeSO4 and KMnO4.
According to the balanced equation, 10 moles of FeSO4 react with 2 moles of KMnO4.
Step 4: Use the stoichiometric ratio to convert moles of FeSO4 to moles of KMnO4.
Multiply the moles of FeSO4 by the ratio of moles of KMnO4 to moles of FeSO4.
Step 5: Convert moles of KMnO4 to milliliters of 0.203 M KMnO4 solution.
To do this, use the molarity (0.203 M) and the definition of molarity:
Molarity = moles/Liter of solution
Divide the number of moles of KMnO4 by the molarity to get the volume of the solution in liters. Then convert liters to milliliters.
Step 6: Calculate the final answer.
Round your answer to the appropriate significant figures.
Remember to show your work step-by-step! Let me know if you need help with any calculations or further explanations.
To set up this problem, we need to use stoichiometry and the given balanced chemical equation.
First, let's determine the molar mass of FeSO4 and KMnO4:
- Molar mass of FeSO4:
FeSO4 = (1 * atomic mass of Fe) + (1 * atomic mass of S) + (4 * atomic mass of O)
- Molar mass of KMnO4:
KMnO4 = (1 * atomic mass of K) + (1 * atomic mass of Mn) + (4 * atomic mass of O)
Next, we need to determine the number of moles of iron(II) sulfate (FeSO4) in 3.64 g:
- Moles of FeSO4 = mass of FeSO4 / molar mass of FeSO4
Then, we can use the balanced chemical equation to determine the ratio of KMnO4 to FeSO4:
- According to the equation:
10 FeSO4(aq) + 2 KMnO4(aq) + 8 H2SO4(aq) → 5 Fe2(SO4)3(aq) + 2 MnSO4(aq) + K2SO4(aq) + 8 H2O(l)
- The stoichiometric ratio of FeSO4 to KMnO4 is 2:10 (from the coefficients).
Finally, we can use the stoichiometric ratio to find the number of moles of KMnO4 required:
- Moles of KMnO4 = moles of FeSO4 * (10/2)
Now, we need to calculate the volume (in milliliters) of 0.203 M KMnO4 that contains the calculated moles of KMnO4:
- Volume (in liters) of 0.203 M KMnO4 = moles of KMnO4 / molarity (M) of KMnO4
- Volume (in milliliters) of 0.203 M KMnO4 = (Volume in liters) * 1000
This will give you the final answer: the number of milliliters of 0.203 M KMnO4 needed to react with 3.64 g of FeSO4.