From the normality of the permanganate solution and the volume of this solution

Question

From the normality of the permanganate solution and the volume of this solution required, calculate the percent by weight of H2O2 in the sample of peroxide titrated. Can the normality of the H2O2 titration by found or is it not needed to find the percent by weight of H2O2. Please show all steps used to find the percent by weight of H2O2. Thank you!

Standardization of the Potassium Permanganate Solution

                                                             Trial 1                                        Trial 2                                                        

Mohr's Salt (1g)                                  __1.0012 g__                          __1.0033 g__             

Titration: (Sulfuric acid solution)

Final (mL):                              __38.1 mL_                            __40.3 mL__

Initial (mL):                             __00.0 mL__                           __00.0 mL__

Total (mL): __38.1 mL__                           __40.3 mL__

Normality:                               _ 6.70 x 10-2 _                         _ 6.35 x 10-2 _            

            Average Normality:    _ 6.525 x 10-2 _          =          _ 6.53 x 10-2 _

            Determination of Peroxide Content of Hydrogen Peroxide content of Hydrogen Peroxide

                                                              Trial 1                                               

H2O2 + Vial (g)                                   _10.7552 g_                                       

Vial (g)                                                _ 9.4122 g_

H2O2 (g)                                 _ 1.3430 g_

Titration: (Sulfuric acid solution)

Final (mL):                              __37.1 mL_   

Initial (mL):                             __00.0 mL__

Total (mL):                  __38.1 mL_

Normality:                               __________   

Percent by Weight:   __________   

Explanation / Answer

FOR THE CALCULATION OF H2O2 PERCENT WE HAVE TO FIRST CALCULATE THE NO. OF MOLES OF KMNO4, FROM AVERAGE NORMALITY OF KMNO4 WE CAN CALCULATE THE moles of kmno4 as,

normality of kmno4=molar mass of kmno4/5,

molarity OF KMNO4 = 5N,

MOLARITY = NORMALITY *5,

AVERAGE NORMALITY OF KMNO4 TRAIL 1= 6.525*10-2, PUTING THIS VALUE IN EQUATION,

MOLARITY= 6.525*10-2 *5 ,

MOLARITY= 0.32625 M KMNO4,

MOLES=MOLARITY * VOLUME IN L,

moles of kmno4= 0.32625* 0.0381 L (38.1 ML/1000=0.0381 L),

MOLES OF KMNO4=0.01243,

2 KMNO4 + 5H2O2 +3 H2SO4 ---------->2 MNSO4 + 5O2+ K2SO4 +8 H2O,

From above equation , for 5 moles of h2o2 one 2 moles of kmno4 are required,

we have 0.01243 moles of kmno4 that means we have 5/2*0.01243 = 0.03107 moles of H2O2,

FROM THAT WE CAN CALCULATE THE MASS OF H2O2 ,

MASS= MOLES* MOL .WT,

MASS = 0.03107* 34.02= 1.0571 GRMAS OF H2O2,

WT PERCENTAGE OF H2O2= WT/WT,   

WT PERCENTAGE OF H2O2= 1.0571/39.1571 = 0.0269*100= 2.69 % OF H2O2 (WT/WT),

also we can calculate the percentage by wt/volume = 1.0571/38.1=0.02774*100=2.77 %, ( VOLUME OF H2O2 USED IS ASSUME HERE AS 38.1 ML)

(you can do the same calculations for the trail 2 as shown above),

now we can calculate the normality and wt percentage from the gms of H2O2 GIVEN,

WT OF H2O2= 1.3430 GRAMS,

MOLES OF H2O2= 1.3430/34.02=0.0394 ,

MOLARITY=MOLES/VOLUME IN L,

=0.0394/0.0381 = 1.0361 M H2O2 (38.1/1000=0.0381),

EQUIVALENT MASS OF H2O2= 34/2=17 GMS,

NORMALITY = MASS /(EQ; MASS* VOLUME ),

=1.3430/(17*0.0381),

NORMALITY OF H2O2 =2.07 N ,

FROM NORMALITY WE CAN CALCULATE THE MOLARITY BUT , WE HAVE CALCULATED THE MOLES OF H2O2 FROM MASS OF H2O2 , HERE MASS OF H2O2 IS GIVEN , THERE FORE NORMALITY DOES NOT NEEDED IN WT PERCENTAGE CALCULATION .

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