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Juliette [100K]
3 years ago
13

How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?

Chemistry
1 answer:
Bumek [7]3 years ago
6 0

Answer:

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

<em>That means, 1 mole of hydroxide reacts per mole of acid</em>

Moles of  20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

<h3>27.6mL of LiOH 0.250M</h3>
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Answer:

d. 60.8 L

Explanation:

Step 1: Given data

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Step 2: Calculate the work (W) done by the system

We will use the following expression.

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W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J ×  1 atm.L/101.325 J = -1.593 atm.L

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First, we will use the following expression.

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ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

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V2 = V1 + ΔV

V1 = V2 - ΔV

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The question pretty  much requires us to find the amount of moles of each compounds based on the number of moles of O given.

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x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

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1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

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