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

Question

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.

Answer 1

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.