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ElenaW [278]
2 years ago
6

Calculate the volume of a 0.225M solution of KOH required to react with 0.215g of acetic acid.

Chemistry
2 answers:
Sidana [21]2 years ago
7 0
Hope this might help ask me if u have any doubts

Fiesta28 [93]2 years ago
7 0

Answer:

V = 0.018 L

Explanation:

Given data:

Mass of acetic acid = 0.215 g

Molarity of KOH = 0.225 M

Volume of KOH required = ?

Solution:

Chemical equation:

CH₃COOH + KOH      →    CH₃COOK + H₂O

Number of moles of acetic acid:

Number of moles = mass/molar mass

Number of moles = 0.215 g / 60.0 g/mol

Number of moles = 0.004 mol

now we will compare the moles of acetic acid and KOH.

                          CH₃COOH       :       KOH

                                  1                :          1

                              0.004           :       0.004

Volume of KOH required:

Molarity = number of moles / volume

0.255 M = 0.004 mol / V

V = 0.004 mol / 0.225 mol/L

V = 0.018 L

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

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For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

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3 years ago
In a 1.0-liter container there are, at equilibrium, 0.10 mole h2, 0.20 mole n2, and 0.40 mole nh3. what is the value of kc for t
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The reaction involved here would be written as:

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The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:

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Answer:

Water uses adhesive forces that allow it to stick to certain surfaces such as glass.

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