You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

Calculate the...

Question

Calculate the percentage of $C r$ in a sample of dichromate ore if $1.0$ g of the sample after fusion is treated with $60$ mL of $0.1 N$ $F e S O_{4} . (N H_{4})_{2} S O_{4}$ and the excess of $F e^{2 +}$ requires $11.2$ mL of $K_{2} C r_{2} O_{7}$ ( $1$ mL of $K_{2} C r_{2} O_{7} = 0.006$ g $F e$ ). (Write your answer as nearest integer.)

Open in App

Solution

The redox changes are as follows:
$6 e^{-} + C r_{2} O_{7}^{2 -} \to 2 C r^{3 +} (n = 6)$
$F e^{2 +} \to F e^{3 +} + e^{-} (n = 1)$

Milliequivalents of $K_{2} C r_{2} O_{7}$ in $1$ mL $=$ Milliequivalents of $F e$ $= \frac{0.006}{56} \times 10^{3} = \frac{6}{56}$

Milliequivalents of $K_{2} C r_{2} O_{7}$ in $11.2$ mL $= \frac{6}{56} \times 11.2 = 1.2$

Milliequivalents of $F e^{2 +}$ left unused $=$ Milliequivalents of $K_{2} C r_{2} O_{7}$ used $= 1.2$

Now, milliequivalents of $F e S O_{4} . (N H_{4})_{2} S O_{4}$ added $= 60 \times 0.1 = 6$

Milliequivalents of $F e S O_{4} . (N H_{4})_{2} S O_{4}$ left unused $= 6 - 1.2 = 4.8$

Milliequivalents of $C r = 4.8$ $(∵ E w o f C r = \frac{52}{3})$

$\frac{W}{\frac{52}{3}} \times 10^{3} = 4.8$

$W_{C r} = \frac{4.8 \times 52}{10^{3} \times 3} = 0.0832$ g

$%$ of $C r = \frac{0.0832}{1.0} \times 100 = 8.32 %$

Also, milliequivalents of $C r_{2} O_{3} = 4.8$ $(∵ E w o f C r_{2} O_{3} = \frac{52 \times 2 + 48}{6} = \frac{156}{6})$

$\frac{W_{C r_{2} O_{3}}}{\frac{152}{6}} \times 10^{3} = 4.8$

$W_{C r_{2} O_{3}} = \frac{4.8 \times 152}{6 \times 1000} = 0.1216$ g

$%$ of $C r_{2} O_{3} = \frac{0.1216}{1.0} \times 100 = 12.16 %$

So, the nearest integer value is $12$ .

Suggest Corrections

Similar questions

Q. If

X %

C r

in a sample of dichromate ore if

1.0 g

of the sample after fusion is treated with

60 m L

0.1 N

F e S O_{4} \cdot (N H_{4})_{2} S O_{4}

and the excess of

F e^{2 +}

requires

11.2 m L

K_{2} C r_{2} O_{7} (1 m L

K_{2} C r_{2} O_{7} = 0.006 g F e)

. Also find

%

C r_{2} O_{3}

in the sample then calculate the value of

100 X

Q. The chromium in a

1.0 g

sample of chromite

(F e C r_{2} O_{4})

was oxidized to

C r^{+ 6}

state by fusion with

N a_{2} O_{2}

. The fused mixture was treated with excess of water and boiled to destroy the excess of peroxide. After acidification, the sample was treated with

50 mL

0.16 M

F e S O_{4}

solution.

6.67 mL

0.05 M

C r_{2} O_{7}^{2 -}

solution was required to oxidize the

F e^{2 +}

ion left unreacted. Determine the mass percentage of chromite in the original sample

[C r = 52, F e = 56]

Q. A

10 g

sample of

C u S

and

C u_{2} S

was treated with

100 m L

1.25 M K_{2} C r_{2} O_{7}

to produce

C r^{3 +}, C u^{2 +}

and

S O_{2}

. The excess oxidant was reacted with

50 m L

F e^{2 +}

solution.

25 m L

of the same

F e^{2 +}

solution required

0.875 M K M n O_{4}

under acidic condition, the volume of

K M n O_{4}

used was

20 m L

. Find the percentage of

C u_{2} S

in the sample.
Given: molar mass of

C u

63.5 g m o l^{- 1}

1.6

g of pyrolusite ore was treated with

50

mL of

0.1 N

oxalic acid and some sulphuric acid. The oxalic acid left in excess was raised to

250

mL in a flask.

25

mL of this solution when titrated with

0.01 N

K M n O_{4}

required

30

mL of the solution. The percentage of pure

M n O_{2}

in the sample is :

1.22

g sample of

{N a}_{2} C O_{3}

and

K_{2} C O_{3}

was dissolved in water to form

100

mL of a solution.

20

mL of this solution required

40

mL of

0.1 N

H C l

for complete neutralization. Calculate the mass of

{N a}_{2} C O_{3}

in the mixture. If another

20 m L

of this solution is treated with excess of

B a {C l}_{2}

, then the mass of pecipitate will be (write the nearest digit just after the decimal point as

0.456 = 5

) :

Join BYJU'S Learning Program

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 FeSO4.(NH4)2SO4 and the excess of Fe2+ requires 11.2 mL of K2Cr2O7 (1 mL of K2Cr2O7=0.006 g Fe). (Write your answer as nearest integer.)