5H2O2 + 2KMnO4<span>+ 3H2SO4 = 5O2 + 2MnSO4 + 8H2O + K2SO4
0,145 moles of KMnO4----------in--------1000ml
x moles of KMnO4---------------in------------46ml
x = 0,00667 moles of KMnO4
according to the reaction:
2 moles of KMnO4------------------5 moles of H2O2
0,00667 moles of KMnO4----------------x
x = 0,01668 moles of H2O2
0,01668 moles of H2O2---------in-----------50ml
x moles of H2O2--------------------in----------1000ml
<u>x = 0,334 mol/L H2O2</u></span>
Answer:
The volume of the stock solution needed is 1L
Explanation:
Step 1:
Data obtained from the question. This include the following:
Concentration of stock solution (C1) = 6M
Volume of stock solution needed (V1) =?
Concentration of diluted solution (C2) = 1M
Volume of diluted solution (V2) = 6L
Step 2:
Determination of the volume of the stock solution needed.
With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
6 x V1 = 1 x 6
Divide both side by 6
V1 = 6/6
V1 = 1L
Therefore, the volume of the stock solution needed is 1L
1 mole of water = 18 grams (you can find this by finding mass of two hydrogen and one oxygen which is (1*2) + 16 = 18)
1.8 grams = 0.1 moles
1 H2O molecule has 10 electrons so 0.1 moles can be computed as:
(6.023*10^23)*(0.1)*10 = 6.023*10^23 electrons
I think the answer would be 1.58 g.
