When N a 2 S 2 O 3 reacts with I 2 to form N a 2 S 4 O 6 and N aI then which statements is correct-A- n-factor for N a 2 S 2 O 3 is oneB- n-factor for I 2 is twoC- 2 moles of N a 2 S 2 O 3 is reacted with 1 mole of I 2D- n-factor for N a 2 S 4 O 6 is one

The correct options are A n factor for nbsp;Na_2S_2O_3 is one nbsp; B n factor for nbsp;I_2 is two C 2 moles of nbsp;Na_2S_2O_3 is reacted with 1 mole o ...

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Byju's Answer

Standard XII

Chemistry

Sodium Thiosulphate

When N a 2 S ...

Question

When $N a_{2} S_{2} O_{3}$ reacts with $I_{2}$ to form $N a_{2} S_{4} O_{6}$ and $N a I$ then which statement(s) is correct?

n-factor for

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

is one

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n-factor for

I_{2}

is two

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2 moles of

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

is reacted with 1 mole of

I_{2}

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n-factor for

N a_{2} S_{4} O_{6}

is one

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Solution

The correct options are
A n-factor for $N a_{2} S_{2} O_{3}$ is one
B n-factor for $I_{2}$ is two
C 2 moles of $N a_{2} S_{2} O_{3}$ is reacted with 1 mole of $I_{2}$
$2 N a_{2} {+ 2 S}_{2} O_{3} + {0 I}_{2} \to N a_{2} {+ 2.5 S}_{4} O_{6} + 2 N a - 1 I$

n-factor of $N a_{2} S_{2} O_{3} = 2 \times (2.5 - 2) = 1$

n-factor of $I_{2} = 2 \times 1 = 2$

n-factor of $N a_{2} S_{4} O_{6} = 4 \times (2.5 - 2) = 2$

Suggest Corrections

Similar questions

Q. When

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

reacts with

I_{2}

to form

N a_{2} S_{4} O_{6}

and

N a I

then which statement(s) is correct?

Q. Identify the oxidant and the reductant in the following reaction respectively:

I_{2} + N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + N a I

Q. In the following reaction,

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

is converted to

N a_{2} S_{4} O_{6}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

The equivalent weight of

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

for this reaction is:
(The molecular weight of

N a_{2} S_{2} O_{3} = M)

Q. x moles of sulfur get disproportionated in a basic medium and form

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

and

N a_{2} S

as:

S + N a O H \to N a_{2} S_{2} O_{3} + N a_{2} S

Sodium thiosulfate reacts with iodine in a basic medium to form

N a I

and

N a_{2} S O_{4}

by the reaction:

N a_{2} S_{2} O_{3} + I_{2} + N a O H \to N a I + N a_{2} S O_{4} + H_{2} O

N a I

is completely precipitated by 2.8 N 100 ml

A g N O_{3}

solution then what is the value of x?

Q. In the following reaction,

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

is converted to

N a_{2} S_{4} O_{6}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

The equivalent weight of

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

for this reaction is:

(The molecular weight of

N a_{2} S_{2} O_{3} = M)

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When Na2S2O3 reacts with I2 to form Na2S4O6 and NaI then which statement(s) is correct?

The correct options are A n-factor for Na2S2O3 is one B n-factor for I2 is two C 2 moles of Na2S2O3 is reacted with 1 mole of I22Na2+2S2O3+0I2→Na2+2.5S4O6+2Na−1I n-factor of Na2S2O3=2×(2.5−2)=1 n-factor of I2=2×1=2 n-factor of Na2S4O6=4×(2.5−2)=2

When $N a_{2} S_{2} O_{3}$ reacts with $I_{2}$ to form $N a_{2} S_{4} O_{6}$ and $N a I$ then which statement(s) is correct?