DAT Achiever question

[RHONDAROBINSON]

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Hello,

Can someone explain these questions to me. I am having trouble understanding it.

(1)What percentage of iron is present in the ore, if a 1.120g sample requires 25.0ml of 0.050M KMnO4 to fully oxidize all of the dissolved Fe+2?
8H +5Fe+2 +MnO4- yields 5Fe+3 + Mn+2 +4H20

The numbers that have +before them is the oxidation number.

(2) Determine the value of (x) if 10.0g of CuSO4.xH2O have been dehydrated to 6.4g of anhydrous CuSO4.


Thanks.

rr

 

[hausee]

I am new to here and this is my forst post, be gentle

anyway 1st question:

every mole of Mn react with 5 mole of Fe, so find out how many mole of Mn reacted with Fe will know the mole of Fe2+ that in the curde iron ore.

mmole of MnO4=25mlx0.05M=1.25mmol
mmole of Fe2+=1.25mmole x 5=6.25mmol
so mass of Fe=6.25mmol x 55.85=349mg
%= 0.349g/1.12g=31.2%

correct me if I am wrong.

thank you

 

[hausee]

2nd question:


mole of 6.4g CuSO4=6.4g/159.61=0.04 mole
mole of (10-6.4) H2O= 0.2 mole

ratio of mole: CuSO4:H2O=0.04:0.2=1:5

so X=5

thank you

 

[RHONDAROBINSON]

Thanks for the help, but I am not understanding where you are getting the number in the question #1.

RR

 

[hausee]

1st question:

every mole of Mn react with 5 mole of Fe, so find out how many mole of Mn reacted with Fe will know the mole of Fe2+ that in the curde iron ore.

mmole of MnO4=25mlx0.05M=1.25mmol <---from 25.0 ml 0.050M KMnO4 solution. 1 ml x (mole/L)=1/1000 L x Mole/L=1/1000 Mole= 1mmole

mmole of Fe2+=1.25mmole x 5=6.25mmol <---every mmole of MnO4 react with 5 mmole of Fe2+

so mass of Fe=6.25mmol x 55.85 g/mole=0.349g <---55.85 is molecular weight of Fe in periodic table.

%= 0.349g/1.12g x 100%=31.2%



thank you