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3 years ago

Potassium hydroxide reacts with hydrochloric acid to form salt and water.

Chemistry

2 answers:

iris [78.8K]3 years ago

8 0

<h3>Answer:</h3>

a. KOH (aq) + HCl (aq) → KCl (s) + H₂O (l)

b. NaOH (aq) + HCl (aq) → NaCl (s) + H₂O (l)

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

Reading a Periodic Table
Writing Compounds
Polyatomic Ions

<u>Aqueous Solutions</u>

States of matter
Solubility Rules

<h3>Explanation:</h3>

We are given "Potassium hydroxide reacts with hydrochloric acid to form salt and water."

From the question, we can figure out 1.) if the reaction exists and 2.) if it does exist, then reaction prediction.

Potassium - K⁺
Hydroxide (PAI) - OH⁻
Potassium Hydroxide - KOH
Hydro - H⁺
Chloric (Chlorine) - Cl⁻
Hydrochloric acid (strong) - HCl

Remembering solubility rules, we have:

<em>All common compounds of Group IA ions are soluble</em>
<em>All common chlorides, bromides, and iodides are mostly soluble (exceptions are Ag⁺, Pb²⁺, Cu⁺, Hg₂²⁺)</em>

Therefore, we can conclude that both potassium hydroxide and hydrochloric are both aqueous solutions:

KOH (aq) + HCl (aq) → ? + ?

We are also given from the problem that the reaction produces a salt and water.

Molecular Formula of water - H₂O

We also use the double replacement reaction to do reaction prediction of the proposed salt:

KOH (aq) + HCl (aq) → KCl (s) + H₂O (l)

Hence, this is our answer to a.).

b.)

We are given that we <em>switch</em> the compound KOH (potassium hydroxide) with sodium hydroxide.

Sodium - Na⁺
Hydroxide (PAI) - OH⁻
Sodium Hydroxide - NaOH

We can use the same steps as a.) to figure out our new reaction. Since we already know that Group IA compounds are soluble, we know that NaOH is soluble:

NaOH (aq) + HCl (aq) → ? + H₂O (l)

Following the same steps, we can simple do a double replacement reaction prediction to find the missing salt:

NaOH (aq) + HCl (aq) → NaCl (s) + H₂O (l)

We see that we would still be able to produce a [different] salt and the reaction does exist/run. And we have our final answer for b.).

While the salts produced (KCl and NaCl) are solids, they <em>are</em> soluble in water. Since we are running only a forward reaction, we know that we are <em>producing</em> a solid and liquid water. Usually, we would probably see these reactions in equilibrium (represented with a ⇆).

This also delves into the concepts of saturation.

finlep [7]3 years ago

6 0

Answer:

word equation

Potassium hydroxide+hydrochloric acid=salt and water.

KoH+HCL_____>KCL+H2O

balanced already

when

sodium hydroxide is used instead of potassium hydroxide

sodium hydroxide+hydrochloric acid=salt and water.

NaoH+HCL_____>NaCl+H2O

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What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

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Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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Explanation:

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