2 g of `FeC_(2)O_(4)` are made to react in acid solution with `0.25 M KMnO_(4)` solution. What volume of `KMnO_(4)` would be required? The be reqruied? The resulting solution is treated with excess of `NH_(4)Cl` solution and `NH_(4)OH` solution. The precipitated `Fe(OH)_(3)` is filtered off, washed and ignited. What is the mass of the product obtained? (Atomic weight of `Fe=56`)

2 g of `FeC_(2)O_(4)` are made to react in acid solution with `0.25 M

1.	2 g of `FeC_(2)O_(4)` are made to react in acid solution with `0.25 M KMnO_(4)` solution. What volume of `KMnO_(4)` would be required? The be reqruied? The resulting solution is treated with excess of `NH_(4)Cl` solution and `NH_(4)OH` solution. The precipitated `Fe(OH)_(3)` is filtered off, washed and ignited. What is the mass of the product obtained? (Atomic weight of `Fe=56`)
Answer» `FeC_(2)O_(4)(144g)-=1+2=3` equivalent `2 g FeC_(2)O_(4)-=(2)/(144)xx3=(1)/(24)` equivalent `0.25 M KMnO_(4)-=1.25N` `(V)/(1000)xx1.25=(1)/(24)impliesV=(1000)/(24xx1.25)=33.33 mL` i.e., volume of `KMnO_(4)` required `=33.33mL` Weight of Fe in the sample is `(56)/(144)xx2=(112)/(144)g`. Since `underset(112g)(2Fe)+3(O)tounderset(160g)(Fe_(2)O_(3))` `(112)/(144)g-=(160)/(112)xx(112)/(144)=1.11g` i.e., the mass of product obtained`=1.11g`

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