Question

Ca(OCl)2 + 2NaCl + 2H2SO4 <=> 2Cl2 + CaSO4 + NaSO4 + 2H2O

[We have to prepare 12 ml of 0.050 M Cl2 (Chlorine water) ]
[We are given "solid Ca(OCl)2" , "0.20 M NaCl" and "0.50 M H2SO4" and "D-I water"]

What is the required quantity of Ca(OCl)2, NaCl and H2SO4 and volume of D-I water to be used??

Answer 1

I tried it myself but I'm not sure it is the right steps...please check

1st - I used 0.050 mol/1 L x .012 L to get .00060 mol Cl2

2nd - To get the Ca(OCl)2 grams I used the 1:2 ratio and figured that there is .00030 mol Ca(OCl)2. I multiplied it with 142.98 g molar mass and got ".043 grams Ca(OCl)2"

3rd - I used the 2:2 ratio and think that there is .00060 NaCl. I multiplied it with 1L/.20 mol and got ".0030 L NaCl"

4th - I used the 2:2 ratio again and used .00060 H2SO4. I multiplied 1L/.50 mol and got ".0012 L H2SO4"

5th - Knowing this I wrote that we needed 7.8 ml to create the 12ml Chlorine solution.

Answer 2

I would modify your procedure.
I agree that you want 0.00060 (don't forget the final 0) mol Cl2.
Note that 1/2 of the 2 Cl2 comes from Ca(OCl)2 and the other half of the 2Cl2 comes from the 2NaCl.
0.00030 x molar mass Ca(OCl)2 = about 43 mg.
For NaCl, 0.0030 mol Cl2/2 = mols NaCl and you can go from there.
For H2SO4 you want twice the mols Ca(OCl)2 or 0.0006 so I think that part is ok, too.

Answer 3

I think everything is good, but the equation is not balanced, Check Na (Sodium).