Calculate the percentage of Cr in a sample of dichromate ore if 1.0 g of the sample after fusion is treated with 60 " mL of " 0.1 N `FeSO_(4).(NH_(4))_(2)SO_(4)` and the excess of `Fe^(2+)` requires `11.2 " mL of " K_(2)Cr_(2)O_(3)` in the sample.

Explanation

`6e^(-)+Cr_(2)O_(7)^(2-)to2Cr^(3+)(n=6)`
`Fe^(2+)toFe^(3+)+e^(-)(n=1)`
`m" Eq of "K_(2)Cr_(2)O_(7)` in 1 mL`=m" Eq of "Fe`
`=(0.006)/(56)xx10^(3)=(6)/(56)`
m" Eq of "`K_(2)Cr_(2)O_(7)` in 11.2 mL`=(6)/(56)xx11.2=1.2` ltBrgt mEw of `Fe^(2+)` left unused`=m" Eq of "K_(2)Cr_(2)O_(7) used=1.2`
Now, `m" Eq of "FeSO_(4).(NH_(4))_(2)SO_(4) added =60xx0.1=6`
`m" Eq of "FesO_(4).(NH_(4))_(2)SO_(4)` left unused`=6-1.2=4.8`
m" Eq of "`Cr=4.8` (`thereforeEw of Cr=(52)/(3))`
`(E)/((52)/(3))xx10^(3)=4.8`
`W_(Cr)=(4.8xx52)/(10^(3)xx3)=0.0832g`
`% of Cr=(0.0832)/(1.0)xx100=8.32%`
Also, m" Eq of "`Cr_(2)O_(3)=4.8`
`(becauseEw of Cr_(2)O_(3)=(52xx2+48)/(6)=(156)/(6))`
`(W_(Cr_(2)O_(3))/((152)/(6))xx10^(3)=4.8`
`W_(Cr_(2)O_(3))=(4.8xx152)/(6xx1000)=0.1216g`
`% of Cr_(2)O_(3)=(0.1216)/(1.0)xx100=12.16%`