how many milliliters of 0.250 M KMnO4 are needed to react with 3.55 grams of iron(II)sulfate, FeSO4? the reaction is as follows:
10FeSO4(aq)+2KMnO4(aq)+8H2SO4(aq)�¨
5Fe2(SO4)3(aq)+2MnSO4(aq)+K2SO4(aq)+8H2O(l)
Well, that's quite a chemistry question! Looks like you're balancing equations and calculating stoichiometry. But let me clown around a little with the answer for you:
To figure out how many milliliters of 0.250 M KMnO4 are needed, we first have to determine the number of moles of FeSO4 in 3.55 grams. Are you aware that FeSO4 stands for "Feeling Stressed Over Chemistry?" Well, let's find out just how stressed we should be!
The molar mass of FeSO4 is 151.91 g/mol. So, using some mathematical magic: 3.55 g / 151.91 g/mol = 0.0234 moles of FeSO4. Hocus pocus!
Now, let's conjure up the mole ratio between KMnO4 and FeSO4 from the balanced equation. From the equation, we can see that 2 moles of KMnO4 react with 10 moles of FeSO4. That's like mixing 2 potions with 10 different magical ingredients, quite a balancing act!
Using this ratio, we can determine how many moles of KMnO4 we need: 0.0234 mol of FeSO4 * (2 mol of KMnO4 / 10 mol of FeSO4) = 0.00468 moles of KMnO4.
Now, let's abracadabra that into milliliters of the KMnO4 solution. Since the concentration is 0.250 M, we know that 0.250 moles of KMnO4 are present in 1 liter (1000 mL) of solution. But we just need 0.00468 moles, so let's use a bit more magical math:
0.00468 mol * (1000 mL / 0.250 mol) = 18.7 mL of 0.250 M KMnO4 solution! Ta-da!
So, it looks like you need approximately 18.7 milliliters of 0.250 M KMnO4 to work your chemical wonders and make the reaction happen. Just be careful not to let the chemistry spills turn you into a clown, too!
To find the number of milliliters of 0.250 M KMnO4 needed, we need to use the concept of stoichiometry. The balanced equation tells us that 10 moles of FeSO4 react with 2 moles of KMnO4.
Step 1: Convert the mass of FeSO4 to moles.
Given mass of FeSO4 = 3.55 grams
Molar mass of FeSO4 = 55.845 + 32.06 + (4 * 16) = 151.908 grams/mol
Number of moles of FeSO4 = 3.55 grams / 151.908 grams/mol = 0.0234 moles
Step 2: Determine the number of moles of KMnO4 required.
According to the balanced equation, 10 moles of FeSO4 react with 2 moles of KMnO4.
So, 0.0234 moles of FeSO4 require (2/10) * 0.0234 moles of KMnO4.
Number of moles of KMnO4 required = (2/10) * 0.0234 moles = 0.00468 moles
Step 3: Calculate the volume of 0.250 M KMnO4 solution using the given molarity.
Molarity of KMnO4 solution = 0.250 M
Number of moles of KMnO4 = 0.00468 moles (from step 2)
Volume of KMnO4 solution (in liters) = Number of moles / Molarity
Volume of KMnO4 solution (in liters) = 0.00468 moles / 0.250 M = 0.01872 liters
Step 4: Convert liters to milliliters.
1 liter = 1000 milliliters
Volume of KMnO4 solution (in milliliters) = 0.01872 liters * 1000 = 18.72 milliliters
Therefore, approximately 18.72 milliliters of 0.250 M KMnO4 solution are needed to react with 3.55 grams of FeSO4.
To find out how many milliliters of 0.250 M KMnO4 are needed to react with 3.55 grams of FeSO4, we need to use stoichiometry and the molar ratios from the balanced equation.
First, let's calculate the number of moles of FeSO4 using its molar mass:
Molar mass of FeSO4 = atomic mass of Fe + atomic mass of S + 4 * atomic mass of O
= (55.845 g/mol) + (32.06 g/mol) + 4 * (16.00 g/mol)
= 55.845 g/mol + 32.06 g/mol + 64.00 g/mol
= 151.905 g/mol
Number of moles of FeSO4 = mass / molar mass
= 3.55 g / 151.905 g/mol
≈ 0.02338 moles
From the balanced equation, we can see that the molar ratio between FeSO4 and KMnO4 is 10:2. This means that for every 10 moles of FeSO4, we need 2 moles of KMnO4.
Now, we can use the molar ratio and the concentration of KMnO4 to find the volume (in liters) of KMnO4 needed to react with the given amount of FeSO4.
Number of moles of KMnO4 = (0.02338 moles FeSO4) * (2 moles KMnO4 / 10 moles FeSO4)
= 0.004676 moles KMnO4
Now, we can use the molarity of KMnO4 to find the volume:
Volume of KMnO4 (in liters) = number of moles of KMnO4 / molarity of KMnO4
= 0.004676 moles / 0.250 M
= 0.0187 L
Finally, convert the volume from liters to milliliters:
Volume of KMnO4 (in milliliters) = 0.0187 L * 1000 mL/L
= 18.7 mL
So, approximately 18.7 milliliters of 0.250 M KMnO4 are needed to react with 3.55 grams of FeSO4.
mols FeSO4 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols FeSO4 to mols KMnO4.
M KMnO4 = mols KMnO4/L KMnO4. You know M and mols, solve for L and convert to mL.