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2 years ago

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

Chemistry

1 answer:

melomori [17]2 years ago

8 0

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

PbCl₂ : Pb(NO₃)₂

1 : 1

0.23 : 0.23

Mass of Pb(NO₃)₂:

Mass = number of moles × molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

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Answer: The pH of acid solution is 4.58

Explanation:

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Putting values in equation 1, we get:

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Final: 0.09206 - 0.04514

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We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

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