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Standard XII

Chemistry

Faraday's Second Law

2Ag NO 3 +Cu→...

Question

$2 A g {N O}_{3} + C u \to C u {({N O}_{3})}_{2} + 2 A g$
What is the percent yield when $0.17 g$ of $A g {N O}_{3}$ in aqueous solution reacts with excess copper to produce $0.08 g$ $A g$ ? (At. mass of $A g = 107 g / m o l$ )

74

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47

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89

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65

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Solution

The correct option is A $74$ %
$2 A g N O_{3} + C u \to C u (N O_{3})_{2} + 2 A g$
Percentage of Ag in $A g N O_{3} = \frac{108 \times 100}{108 + 14 + 48} = \frac{108}{170} \times 100 = \frac{1080}{17} = 63.52 %$
So, amount of Ag produced $= \frac{63.52}{100} \times 0.17 = 0.108 g$
$%$ Yield $= \frac{0.08}{0.108} \times 100 = 74 %$

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Similar questions

What coefficient of $A g N O_{3}$ is required to balance the following equation?

$C u + A g N O_{3} \to C u (N O_{3})_{2} + A g$

Q. An aqueous solution containing one mole per litre of each

C u (N O_{3})_{2}, A g N O_{3}, H g_{2} (N O_{3})_{2}

and

M g (N O_{3})_{2}

, is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are

A g / A g^{+} = + 0.80, 2 H g / H g_{2}^{2 +} = + 0.79, C u / C u^{2 +} = + 0.34, M g / M g^{2 +} = - 2.37

with increasing voltage, the sequence of deposition of metals on the cathode will be

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. Assertion: It is safe to stir 1 M

A g N O_{3}

solution with a copper spoon.
Reason:

E_{A g^{\oplus} | A g}^{⊖} = 0.80 V

E_{C u^{2 +} | C u}^{⊖} = 0.34 V

A solution containing 1 mol per litre of each $C u (N O_{3})_{2}, A g N O_{3},$ and $H g_{2} (N O_{3})_{2}$ is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts (reduction potential) are
$A g | A g^{⨁} = + 0.80, 2 H g | H g_{2}^{2 +} = + 0.79$
$C u | C u^{2 +} = + 0.34, M g | M g^{2 +} = - 2.37$
with increasing voltage, the sequence of deposition of metals at the cathode will be
(IIT-JEE, 1984)

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2AgNO3+Cu→Cu(NO3)2+2AgWhat is the percent yield when 0.17 g of AgNO3 in aqueous solution reacts with excess copper to produce 0.08 g Ag? (At. mass of Ag=107 g/mol)

The correct option is A 74%2AgNO3 + Cu→Cu(NO3)2 + 2AgPercentage of Ag in AgNO3=108×100108+14+48=108170×100=108017=63.52%So, amount of Ag produced =63.52100×0.17=0.108 g% Yield =0.080.108×100=74%

$2 A g {N O}_{3} + C u \to C u {({N O}_{3})}_{2} + 2 A g$
What is the percent yield when $0.17 g$ of $A g {N O}_{3}$ in aqueous solution reacts with excess copper to produce $0.08 g$ $A g$ ? (At. mass of $A g = 107 g / m o l$ )