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xeze [42]
3 years ago
14

PLS HELP

Chemistry
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

Percent yield of silver obtained = 78.18% (Approx.)

Explanation:

Given:

Amount of copper = 50 gram

Amount of silver nitrate = 300 gram

Amount of silver = 149 gram

Find:

Percent yield of silver obtained

Computation:

Percentage of Silver in silver nitrate =  108[100/(108+14+48)]

Percentage of Silver in silver nitrate = 108/[100/170]

Percentage of Silver in silver nitrate = 63.53% (Approx.)

Amount of Silver produced = [63.53%][300]

Amount of Silver produced = 190.59 gram

Percent yield of silver obtained = [Amount of silver / Amount of silver produced]100

Percent yield of silver obtained = [149/190.59]100

Percent yield of silver obtained = 78.18% (Approx.)

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Tanzania [10]
Molar mass:

KCl = 74.55 g/mol

KClO3 = 122. 55 g/mol

  <span>Calculation of the mass of KClO3 :</span>

<span>2 KClO3 =  2 KCl + 3 O2</span>

2* 122.55 g KClO3 ------------------ 2 * 74.55 g KCl
mass KClO3 ?? --------------------- 25.6 g KCl

mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55

mass KClO3 = 6274.56 / 149.1

mass = 42.082 g of  KClO3

Therefore:

1 mole KClO3 ---------------------- 122.55 g
?? moles KClO3 ------------------- 42.082 g

moles KClO3 = 42.082 * 1 / 122.55

moles KClO3 = 42.082 / 122.55

=> 0.343 moles of KClO3


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