Reactions involving gold have been of particular interest to alchemists- Consider the following reactions- AuOH3 - 4 HCl - HAuCl4 - 3 H2O- H - -28 kcalAuOH3 - 4 HBr - HAuBr4 - 3 H2O- H - -36-8 kcalIn an experiment- there was an absorption of 0-44 kcal when one mole of HAuBr4 was mixed with 4 moles of HCl- Then- the fraction of HAuBr4 converted into HAuCl4 percentage conversion

Au(OH)_3 + 4 HCl → HAuCl_4 + 3 H_2O, H = 28 kcal HAu(Br)_4 + 3 H_2O → Au(OH)_3 + 4 HBr, H = +36.8 kcal ___________________________________________________ ...

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Byju's Answer

Standard XII

Chemistry

Heat of Reaction

Reactions inv...

Question

Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,
$A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O, △ H = - 28 k c a l$
$A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O, △ H = - 36.8 k c a l$
In an experiment, there was an absorption of 0.44 kcal when one mole of $H A u B r_{4}$ was mixed with 4 moles of HCl. Then, the fraction of $H A u B r_{4}$ converted into $H A u C l_{4}$ (percentage conversion)

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Solution

The correct option is A 5%
$A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O, △ H = - 28 k c a l$
$H A u (B r)_{4} + 3 H_{2} O \to A u (O H)_{3} + 4 H B r, △ H = + 36.8 k c a l$
___________________________________________________
$H A u B r_{4} + 4 H C l \to H A u C l_{4} + 4 H B r, △ H = + 8.8 k c a l$

When 100% conversion occurs, heat absorbed = 8.8 kcal
% conversion $= \frac{0.44}{8.8} \times 100 = 5$

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Similar questions

Q. Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,

A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O, △ H = - 28 k c a l

A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O, △ H = - 36.8 k c a l

In an experiment, there was an absorption of 0.44 kcal when one mole of

H A u B r_{4}

was mixed with 4 moles of HCl. Then, the fraction of

H A u B r_{4}

converted into

H A u C l_{4}

(percentage conversion)

Q. Reaction involving gold have been of particular interest to a chemist. Consider the following reactions,

A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O

Δ H = - 28 k C a l

A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O

Δ H = - 36.8 k C a l

In an experiment there was an absorption of

0.44 k C a l

when one mole of

H A u B r_{4}

was mixed with

4 m o l e s

H C l

. What is the percentage conversion of

H A u B r_{4}

into

H A u C l_{4}

Q. 4 moles of sodium carbonate is mixed with 6 moles of

H C l

, which gives sodium chloride by the following reaction.

N a_{2} C O_{3} + 2 H C l \to 2 N a C l + C O_{2} + H_{2} O

Calculate the amount of excess reagent in moles.

Q. If 2 moles of NaOH reacts with 4 moles of HCl to give sodium chloride and then the produced sodium chloride reacts with 3 moles of concentrated sulphuric acid to give sodium sulphate, find the total moles of sodium sulphate produced. The concerned reactions are

N a O H + H C l (d i l u t e d) \to N a C l + H_{2} O

2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l

Q. Calculate the mass of

S e O_{3}^{2 -}

in solution on the basis of following data.

20

mL of

\frac{M}{60}

solution of

K B r O_{3}

was added to a definite volume of

S e O_{3}^{2 -}

solution. The bromine evolved was removed by boiling and excess of

K B r O_{3}

was back titrated with

5.1

mL of

\frac{M}{25}

solution of

N a A s O_{2}

. The reactions are given below:

(a)

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ S e O_{4}^{2 -} + B r_{2} + H_{2} O

(b)

B r O_{3}^{-} + A s O_{2}^{-} + H_{2} O ⟶ B r^{-} + A s O_{4}^{3 -} + H^{+}

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The correct option is A 5%Au(OH)3+4 HCl→HAuCl4+3 H2O, △H=−28 kcal HAu(Br)4+3 H2O→Au(OH)3+4 HBr, △H=+36.8 kcal ___________________________________________________ HAuBr4+4 HCl→HAuCl4+4 HBr, △H=+8.8kcal When 100% conversion occurs, heat absorbed = 8.8 kcal % conversion =0.448.8×100=5