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zaharov [31]
3 years ago
11

CO2+2H2O=H3O^++HCO3^-

Chemistry
1 answer:
bekas [8.4K]3 years ago
5 0

<u>Answer:</u>

<em>Bronsted Lowry Theory</em>

<em>CO_2 is the acid </em>

<em>H_2 O is the base </em>

<em>H_3 O^+ is the conjugate acid and </em>

<em>HCO_3^- is the conjugate base</em>

<u>Explanation:</u>

Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

A base is a substance which can accept one or more protons

Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.  

Thus Hydrogen atom has no Charge and it is neutral.

When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

Thus Hydrogen ion has a positive charge.  

Hydrogen ion is also called as a proton since it has only 1 proton in it.

Hydrogen ion in water that is,  

H^++H_2O\Rightarrow H_3O^+

H_3O^+ is called as Hydronium ion.

Acid loses H+ (proton) to form conjugate Base

Base gains H+ (proton) to form conjugate Acid

For example  

Let us consider the example given in the question

CO_2+2H_2 O>H_3 O^+   +HCO_3^- can be written as (removed 1 H_2 O from both the sides )

CO_2+H_2 O>H^++HCO_3^- or  

CO_2+H_2 O  > H_2 CO_3 reversing the equation  

H_2 CO_3+H_2 O>H_3 O^+  +HCO_3^-

H_2 CO_3 is the acid which donates H^+ to form HCO_3^-

H_2 O is the base which gains H^+ to form H_3 O^+

HCO_3^- is the conjugate base and H_3 O^+ is the conjugate acid

So

CO_2 is the acid

H_2 O is the base

H_3 O^+ is the conjugate acid and

HCO_3^- is the conjugate base

(Answer)  

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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 :

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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}}

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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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