0-56 g of limestone was treated with oxalic acid to give CaC2O4- The precipitate decolourised 50 mL of M25 KMnO4 in acidic medium- If X percentage of CaO is present in limestone- then the value of X is-

The correct option is B 50 LimestoneCaCO_3CaC_2O_4MnO^⊝_4→ Decolourises mEq of CaCO_3 = mEq of CaC_2O_4 =mEq of MnO^⊝_4 mEq of Ca ...

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Byju's Answer

Standard XII

Chemistry

Molarity

0.56 g of lim...

Question

$0.56$ g of limestone was treated with oxalic acid to give $C a C_{2} O_{4}$ . The precipitate decolourised $50 m L$ of $\frac{M}{25}$ $K M n O_{4}$ in acidic medium. If $X$ percentage of $C a O$ is present in limestone, then the value of $X$ is:

50

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40

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70

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Solution

The correct option is B $50$
$C a C O_{3} L i m e s t o n e O x a l i c - --- \to a c i d C a C_{2} O_{4} M n O_{4}^{⊝} \to D e c o l o u r i s e s$
mEq of $C a C O_{3}$ = mEq of $C a C_{2} O_{4}$ =mEq of $M n O_{4}^{⊝}$
mEq of $C a C O_{3}$ =mEq of $C a O$
(Since $C a O$ is present in $C a C O_{3}$ )
mEq of $C a O$ = mEq of $M n O_{4}^{⊝}$
$\frac{W}{\frac{56}{2}} \times 10^{3} = 50 \times \frac{1}{25} \times 5$ ( $n$ factor)
$W_{C a O} = 0.28 g$
$%$ of $C a O$ in lime stone $= \frac{0.28 \times 100}{0.56} = 50 %$

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Similar questions

0.56 g

of limestone was treated with oxalic acid to give

C a C_{2} O_{4}

. The precipitate decolorised

45 m L

0.2 N K M n O_{4}

in acid solution. Calculate the % of

C a O

in limestone.

0.518

g sample of limestone is dissolved and calcium is precipitated as calcium oxalate

(C a C_{2} O_{4})

. After filtration and washing the precipitate, it requires

40

mL of

0.05 M K M n O_{4}

solution acidified with sulphuric acid to titrate it. What is the percentage of

C a O

in the limestone? (as nearest integer)

3.2 g

of pyrolusite

(M n O_{2})

was treated with

50 m L

0.5 M

oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to

250 m L

in a flask.

25 m L

of this solution when treated with

0.02 M K M n O_{4}

required

32 m L

of the solution. Find the percentage of

M n O_{2}

in the sample.
Given: Atomic mass of

M n

55 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 :

Q. A

0.518

g sample of limestone is dissolved in

H C l

and the the calcium is precipitated as

C a C_{2} O_{4}

. After filtering and washing the precipitate, it requires

40.0

mL of

0.250 N K M n O_{4}

solution acidified with

H_{2} S O_{4}

to titrate it as. The percentage of

C a O

in the sample is :

M n O_{4}^{-} + H^{+} + C_{2} O_{4}^{2 -} ⟶ M n^{2 +} + C O_{2} + 2 H_{2} O

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0.56 g of limestone was treated with oxalic acid to give CaC2O4. The precipitate decolourised 50mL of M25 KMnO4 in acidic medium. If X percentage of CaO is present in limestone, then the value of X is:

The correct option is B 50CaCO3LimestoneOxalic−−−−→acidCaC2O4MnO⊝4→DecolourisesmEq of CaCO3= mEq of CaC2O4=mEq of MnO⊝4mEq of CaCO3=mEq of CaO(Since CaO is present in CaCO3)mEq of CaO= mEq of MnO⊝4W562×103=50×125×5 (n factor)WCaO=0.28g% of CaO in lime stone =0.28×1000.56=50%

$0.56$ g of limestone was treated with oxalic acid to give $C a C_{2} O_{4}$ . The precipitate decolourised $50 m L$ of $\frac{M}{25}$ $K M n O_{4}$ in acidic medium. If $X$ percentage of $C a O$ is present in limestone, then the value of $X$ is: