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Standard XII

Chemistry

Introduction to Oxidation and Reduction

On heating, a...

Question

On heating, a sample of $N a C l O_{3}$ , it gets converted to $N a C l$ with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as $A g C l$ . The mass of $A g C l$ (in g) obtained will be :
(Given molar mass of $A g C l = 143.5 g m o l^{- 1}$ )

0.35

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0.54

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0.41

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0.48

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Solution

The correct option is D $0.48$
The molar mass of $O_{2} = 32 g / m o l$
$0.16g of oxygen = \frac{0.16 g}{32 g / m o l} = 0.005 m o l$
$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$N a C l + A g N O_{3} \to A g C l + N a N O_{3}$

3 moles of $O_{2}$ =2 moles of $N a C l$ =2 moles of $A g C l$ .
$0.005 moles of O_{2} = 0.005 \times \frac{2}{3} moles of AgCl$
$Molar mass of A g C l = 143.5 g m o l^{- 1}$
The mass of $A g C l$ (in g) obtained will be
$143.5 g / m o l^{- 1} \times 0.005 \times \frac{2}{3} m o l = 0.48 g$

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Similar questions

Q. The mass of

A g C l

obtained by the reaction of

169.8 g

A g N O_{3}

with

29.2 g

N a C l

according to following equation is

27.2 g

A g N O_{3} + N a C l \to A g C l + N a N O_{3} (not balanced)

Calculate the percentage yield.

(

Molar mass of

A g N O_{3}

169.8 g / m o l

, molar mass of

N a C l

58.4 g / m o l

and molar mass of

A g C l

143.3 g / m o l)

Q. A

100 m L

solution containing

A g N O_{3}

was treated with excess

N a C l

to completely precipitate the silver as

A g C l

. If

5.7 g A g C l

was obtained, what was the concentration of

A g^{+}

in the original solution? (At. mass of

A g = 108 g / m o l

)

Q. A

1.0

g sample of a pure organic compound containing chlorine is fused with

N a_{2} O_{2}

to convert chlorine to

N a C l

. The sample is then dissolved in water and the chloride precipitated out on adding

A g N O_{3}

giving

1.96

g of

A g C l

. If the molecular weight of organic compound is

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g/mol, how many chlorine atoms does each molecule contain?

Q. The mass of

A g C l

obtained by the reaction of

169.8 g

A g N O_{3}

with

29.2 g

N a C l

according to following equation is

27.2 g

A g N O_{3} + N a C l \to A g C l + N a N O_{3} (not balanced)

Calculate the percentage yield.

(

Molar mass of

A g N O_{3}

169.8 g / m o l

, molar mass of

N a C l

58.4 g / m o l

and molar mass of

A g C l

143.3 g / m o l)

Q. What weight of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of

A g N O_{3}

?
(Molar mass of Na = 23 g/mol, Cl = 35.5 g/mol, Ag = 108 g/mol)

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On heating, a sample of NaClO3, it gets converted to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be : (Given molar mass of AgCl=143.5 g mol−1)

On heating, a sample of $N a C l O_{3}$ , it gets converted to $N a C l$ with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as $A g C l$ . The mass of $A g C l$ (in g) obtained will be :
(Given molar mass of $A g C l = 143.5 g m o l^{- 1}$ )