#200# ml solution of #H_2O_2# completely oxidises 125g # Fe^+#. Find out the molarity and volume strength of #H_2O_2#?

1 Answer
Doc048
May 28, 2017

#[H_2O_2] = 5.6M# and Vol Stgth (g/L) = 190 g/L

Explanation:

In an acid solution
#H_2O_2 + 2Fe^(2+) + 2H^+# => #2Fe^(3+) + 2H_2O#

From the reaction ratios between #H_2O_2# and #Fe^(2+)#

2 moles #H_2O_2# = 1 mole #Fe^(2+)# = 1 mole #Fe^(3+)#

#2((0.20L)#[#H_2O_2#] = #((125g(Fe^(3+)))/((56"g/(mol))Fe^(3+))))# = #2.23molFe^(3+)#

#[H_2O_2] = ((2.23)/((2)(0.20)))MolarH_2O_2 = 5.60M "in" H_2O_2#

#Vol. "Strength" (g/L)# = #(((5.6"mol"H_2O_2))/L)(0.200L) = 1.12molH_2O_2# = #38 gH_2O_2 "in" 200ml# => #0.190g(H_2O_2)/(ml)#=> #((190gH_2O_2))/L)#

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