Calculate the mass of K2Cr2O7 required to produce 252 g I2 from KI solution- K2Cr2O7-KI- Cr3-I2Molar mass of K2Cr2O7 is 294 g mol-1

K_2+6Cr_2O_7+K 1I→+3Cr^3++0I_2 n_f=(|Change in O.S.|)×Number of atoms n_f_K_2Cr_2O_7=|+3 (+6)|× 2=6 n_f_I_2=|0 ( 1)|× 2=2 From the law of equivalence, Equival ...

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Byju's Answer

Standard XII

Chemistry

Valency and Oxidation State

Calculate the...

Question

Calculate the mass of $K_{2} C r_{2} O_{7}$ required to produce $252 g I_{2}$ from $K I$ solution.
$K_{2} C r_{2} O_{7} + K I \to C r^{3 +} + I_{2}$

(Molar mass of $K_{2} C r_{2} O_{7}$ is $294 g m o l^{- 1}$ )

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Solution

$K_{2} + 6 C r_{2} O_{7} + K - 1 I \to + 3 C r^{3 +} + {0 I}_{2}$

$n_{f} = (| Change in O.S. |) \times Number of atoms$
$n_{f_{K_{2} C r_{2} O_{7}}} = | + 3 - (+ 6) | \times 2 = 6$
$n_{f_{I_{2}}} = | 0 - (- 1) | \times 2 = 2$

From the law of equivalence,
$Equivalents of K_{2} C r_{2} O_{7} = Equivalents of I_{2}$
$\frac{W_{K_{2} C r_{2} O_{7}}}{294} \times 6 = \frac{252}{252} \times 2$
$∴ W_{K_{2} C r_{2} O_{7}} \approx 98 g$

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Similar questions

Q. Calculate the mass of

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

in grams required to produce

254 g I_{2}

from

K I

solution.

K_{2} C r_{2} O_{7} + K I \to C r^{3 +} + I_{2}

(Molar mass of

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

294 g m o l^{- 1}

)

0.5

g mixture of

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

and

K M n O_{4}

is treated with excess of

K I

in acidic medium.

I_{2}

liberated requires

100

mL of

0.15 N

N a_{2} S_{2} O_{3}

solution for titration.

%

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

and

%

K M n O_{4}

are:

0.5

g mixture of

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

and

K M n O_{4}

was treated with excess of

K I

in acidic medium. Iodine liberated required

150 c m^{3}

0.10 N

solution of thiosulphate solution for titration. Find the percentage of

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

in the mixture.

Q. How many grams of

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

is required to oxidise

12.41

g of

F e S O_{4}

F e_{2} (S O_{4})_{3}

in an acidic medium?

(K_{2} C r_{2} O_{7} = 294; F e S O_{4} = 152)

Q. Solve this:

47. Ratio of coefficients of K₂Cr₂O₇ and KI in the reaction, K₂Cr₂O₇+KI+H₂SO₄

\to

Cr₂(SO₄)₃+H₂O+I₂ is:
(a) 3:1 (b) 1:3 (c) 6:1 (d) 1:6

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Calculate the mass of K2Cr2O7 required to produce 252 g I2 from KI solution. K2Cr2O7+KI→Cr3++I2 (Molar mass of K2Cr2O7 is 294 g mol−1)

K2+6Cr2O7+K−1I→+3Cr3++0I2 nf=(|Change in O.S.|)×Number of atoms nfK2Cr2O7=|+3−(+6)|×2=6 nfI2=|0−(−1)|×2=2 From the law of equivalence, Equivalents of K2Cr2O7=Equivalents of I2 WK2Cr2O7294×6=252252×2 ∴ WK2Cr2O7≈98 g

Calculate the mass of $K_{2} C r_{2} O_{7}$ required to produce $252 g I_{2}$ from $K I$ solution.
$K_{2} C r_{2} O_{7} + K I \to C r^{3 +} + I_{2}$

(Molar mass of $K_{2} C r_{2} O_{7}$ is $294 g m o l^{- 1}$ )