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2 years ago

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Write ionic equations for the following: HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

1 answer:

vodomira [7]2 years ago

4 0

Answer:

HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)

Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:

moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl

moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH

You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC

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g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

RUDIKE [14]

<u>Answer:</u>

<u>For a:</u> Lead iodide is a yellow precipitate.

<u>For b:</u> Barium sulfate is a white precipitate.

<u>For c:</u> Ferric hydroxide is a brown precipitate.

<u>For d:</u> Copper (II) hydroxide is a blue precipitate.

<u>Explanation:</u>

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

<u>For (a):</u>

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

<u>For b:</u>

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

<u>For c:</u>

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

<u>For d:</u>

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

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Answer:

Hyu

Explanation:

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Lerok [7]

Answer:

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Explanation:

Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.

The amount of NaCl required is calculated as follows:

(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol

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