Question

calculate the volume of a 0.206M HI solution that needed to reduce 22.5mL of a 0.374 KMnO4 solution ? help plzzzzz

10HI + 2KMnO4 + 3H2SO4 ------> 5i2 + 2MnSO4 = 8H2O

Answer 1

is there any chemical equation listed in your question?

Answer 2

Ooh sorry i forget it !!

10HI + 2KMnO4 + 3H2SO4 ------> 5i2 + 2MnSO4 = 8H2O

Answer 3

its okay :)
Calculate the number of moles=Concentration(mol/dm^3)*Volume(dm^3)
We change ml to dm^3 by dividing by 1000
Number of moles KMnO4=0.374*22.5/1000=0.008415
10HI + 2KMnO4 + 3H2SO4 ------> 5i2 + 2MnSO4 = 8H2O
10 moles of HI are required to react with 2 moles of KMnO4 what if 0.008415 were used of KMnO4 how many moles are required by HI.
Calculate the number of moles=Concentration(mol/dm^3)*Volume(dm^3)
Rearrange your equation to find Volume
Volume=Number of Moles/Concentrate
the answer you will be getting is in dm^3 so make sure you multiply it by 1000

Answer 4

M(HI)*V(HI)/M(KMnO4)*V(KMnO4)=HI/KMnO4
M(HI) represent the concentration of HI
V(HI) represent the volume of HI
M(KMnO4) represent the concentration of KMnO4
M(KMnO4) represent the volume of KMnO4
HI/KMnO4=Mole ratio which is 10/2 according to your chemical equation.

Answer 5

thnxxx bro ^^

Answer 6

your most welcome :)