10L of hard water required 5.6g of lime for removing hardness. Hence temporary hardness in ppm of cac03.

The answer 1000

I think ppm= 1mg/1L
10L x 100g/mol cac03 = 1000

Who check it for me step by step
Thank all tutors for your help.

How many ml of a 0.05M kmno4 solution are required to oxide 2g of fes04 in a dilute solution.

To find out how many ml of a 0.05M KMnO4 solution are required to oxidize 2g of FeSO4, you can follow these steps:

1. Determine the molar mass of FeSO4:
- Fe: 55.85 g/mol
- S: 32.07 g/mol
- O4: (16 * 4) g/mol = 64 g/mol
Therefore, the molar mass of FeSO4 is 55.85 + 32.07 + 64 = 151.92 g/mol.

2. Calculate the number of moles of FeSO4:
Moles = Mass / Molar mass
Moles = 2g / 151.92 g/mol ≈ 0.0131 mol

3. Determine the stoichiometry between FeSO4 and KMnO4:
The balanced equation for the reaction is:
10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + 8H2O + K2SO4
From the equation, we can see that it takes 2 moles of KMnO4 to react with 10 moles of FeSO4.

4. Calculate the number of moles of KMnO4 required:
Moles of KMnO4 = (2 moles KMnO4 / 10 moles FeSO4) * 0.0131 mol FeSO4
Moles of KMnO4 ≈ 0.00262 mol

5. Use the formula c = n / v to calculate the required volume of the KMnO4 solution:
Concentration (c) = 0.05 M
Moles (n) = 0.00262 mol
Rearrange the formula to solve for volume (v): v = n / c
v = 0.00262 mol / 0.05 M ≈ 0.0524 L

6. Convert the volume to milliliters (ml):
1 L = 1000 ml
v = 0.0524 L * 1000 ml/L ≈ 52.4 ml

So, approximately 52.4 ml of the 0.05M KMnO4 solution are required to oxidize 2g of FeSO4.

To find out how many mL of a 0.05M KMnO4 solution are required to oxidize 2g of FeSO4 in a dilute solution, you need to use the concept of stoichiometry.

Step 1: Determine the molar mass of FeSO4:
The molar mass of FeSO4 can be calculated by adding up the atomic masses of its constituent atoms:
Fe: 55.845 g/mol
S: 32.06 g/mol
O: 16.00 g/mol (each oxygen atom)
4 molecules of O: 4 x 16.00 g/mol = 64.00 g/mol

Adding them up gives the molar mass of FeSO4:
55.845 g/mol + 32.06 g/mol + 64.00 g/mol = 151.905 g/mol

Step 2: Determine the moles of FeSO4:
Given the mass of FeSO4 is 2g, we can convert this to moles using the molar mass calculated in step 1:
2g / 151.905 g/mol = 0.0132 mol

Step 3: Write the balanced chemical equation:
The balanced chemical equation for the reaction between KMnO4 and FeSO4 is:
10 FeSO4 + 2 KMnO4 + 8 H2SO4 -> 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O

Step 4: Determine the stoichiometry:
From the balanced chemical equation, we can see that 10 moles of FeSO4 react with 2 moles of KMnO4. Therefore, the ratio is 10:2 or 5:1.

Step 5: Calculate the moles of KMnO4 required:
Since we have determined the moles of FeSO4 in step 2, we can use the stoichiometric ratio to determine the moles of KMnO4 required:
0.0132 mol FeSO4 x (1 mol KMnO4 / 5 mol FeSO4) = 0.00264 mol KMnO4

Step 6: Calculate the volume of KMnO4 solution required:
Using the molarity (M) given as 0.05M, we can convert the moles to volume using the equation:
Molarity (M) = moles / volume (L)

Rearranging the equation, we get:
Volume (L) = moles / molarity (M)
Volume (L) = 0.00264 mol / 0.05 mol/L = 0.0528 L

Step 7: Convert the volume to mL:
To convert from L to mL, we multiply the volume by 1000:
0.0528 L x 1000 mL/L = 52.8 mL

Therefore, approximately 52.8 mL of a 0.05M KMnO4 solution is required to oxidize 2g of FeSO4 in a dilute solution.

1. Why do you think 1000 ppm is the correct answer for #1?

2. Balance the redox part. The other parts of the equation (H2O, H^+, SO4^2-, etc) do not need to be balanced.

a. KMnO4 + 5FeSO4 ==> Mn^2+ + 5Fe^3+
(Mn changes from +7 on the left to +2 on the right for the gain of 5 electrons. Fe is +2 on the left and +3 on the right for the loss of 1 electron so it must be multiplied by 5 to obtain 5 electrons lost.)
b. Convert 2g FeSO4 to mols. mols FeSO4 = grams/molar mass = 2/about 152 = about 0.0132
c. Convert 0.0132 mols FeSO4 to mols KMnO4 using the coefficients in the balanced equation.
0.0132 mols FeSO4 x (1 mol KMnO4/5 mols FeSO4) = 0.0132 x (1/5) = about 0.00263
d. M KMnO4 = mols KMnO4/L KMnO4 or rearrange to
L KMnO4 = mols KMnO4/M KMnO4 = about 0.00263 mols/0.05M = about 0.0527 L = about 52.7 mL.

You should go through the calculations and refine them. I've estimated here and there but the 52.7 mL is close.
If you have further questions please refer to the step you don't understand and clearly explain what you don't understand about that step.