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Zigmanuir [339]
3 years ago
6

Write the acid-base reaction that occurs when an aqueous solution of HCl is added to an aqueous solution of NaOH. (Use the lowes

t possible coefficients. Omit states-of-matter in your answer.)
For Data Table B, the pH of baking soda (sodium bicarbonate) was measured. Answer the following questions based on your observations. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
(a) Write the dissolution reaction for solid sodium bicarbonate below.
Once the ionic solid has dissolved, theamphiproticanion that is formed is able to react as an acid or as a base with water. (For questions b and c, omit states-of-matter in your answer.)
(b) Write the acid-base reaction where hydrogen carbonate is the acid and water is the base.
(c) Write the acid-base reaction where water is the acid and hydrogen carbonate is the base.
Chemistry
2 answers:
Sloan [31]3 years ago
4 0
H3O+(aq) + OH-(aq) --> 2H2O (l)

NaHCO3(s) --> NaH 2+ (aq) + CO3 2- (aq)

NaH 2+ (aq) + H2O (l) --> Na+ (aq) + H3O+ (aq)

H2O (l) + CO3 2- (aq) --> OH- (aq) + HCO3- (aq)

(I'm not completely sure if I did the third question right) I'm sorry if I got it wrong
Pepsi [2]3 years ago
3 0

<u>Answer:</u> The chemical equations are written below.

<u>Explantaion:</u>

An acid-base reaction is known as neutralization reaction.

Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce a salt and water molecule.

The chemical equation for the reaction of HCl and NaOH follows:

HCl+NaOH\rightarrow NaCl+H_2O

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of sodium chloride and 1 mole of water molecule.

  • <u>For a:</u>

The chemical equation for the dissolution of solid sodium bicarbonate in water follows:

NaHCO_3(s)+H_2O(l)\rightarrow Na^+(aq.)+HCO_3^-(aq.)

By Stoichiomtery of the reaction:

1 mole of solid sodium bicarbonate reacts with 1 mole of water, to produce 1 mole of aqueous solution of sodium ions and 1 mole of aqueous solution of hydrogen carbonate ions.

  • <u>For b:</u>

An acid is a substance which looses hydrogen ion when dissolved in water and a base is defined as the substance which accepts hydrogen ion.

The chemical equation where hydrogen carbonate is an acid and reacts with water follows:

HCO_3^-+H_2O\rightarrow CO_3^{2-}+H_3O^+

By Stoichiometry of the reaction:

1 mole of hydrogen carbonate ion reacts with water to produce carbonate ions and hydronium ions.

  • <u>For c:</u>

The chemical equation where hydrogen carbonate is an base and reacts with water follows:

HCO_3^-+H_2O\rightarrow H_2CO_3+OH^-

By Stoichiometry of the reaction:

1 mole of hydrogen carbonate ion reacts with water to produce carbonic acid and hydroxide kions.

Hence, the chemical equations are written above.

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 a. Moles of gold = 35.12/197 = 0.18 moles. b. number of gold molecules = moles x 6.02 x 10^23 = 0.18 x 6.02 x 10^23= 1.08 x 10^23
8 0
4 years ago
26.9 g of solid lithium is combined with 20.0 grams of nitrogen gas. .
SOVA2 [1]

Answer:

The theoretical yield of

Li

3

N

is

20.9 g

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Explanation:

Balanced Equation

6Li(s)

+

N

2

(

g

)

→

2Li

3

N(s)

In order to determine the theoretical yield, we must first find the limiting reactant (reagent), which will determine the greatest possible amount of product that can be produced.

Molar Masses

Li

:

6.941 g/mol

N

2

:

(

2

×

14.007

g/mol

)

=

28.014 g/mol

Li

3

N

:

(

3

×

6.941

g/mol Li

)

+

(

1

×

14.007

g/mol N

)

=

34.83 g/mol Li

3

N

Limiting Reactant

Divide the mass of each reactant by its molar mass, then multiply times the mole ratio from the balanced equation with the product on top and the reactant on bottom, then multiply times the molar mass of

Li

3

N

.

Lithium

12.5

g Li

×

1

mol Li

6.941

g Li

×

2

mol Li

3

N

6

mol Li

×

34.83

g Li

3

N

1

mol Li

3

N

=

20.9 g Li

3

N

Nitrogen Gas

34.1

g N

2

×

1

mol N

2

28.014

g N

2

×

2

mol Li

3

N

1

mol N

2

×

34.83

g Li

3

N

1

mol Li

3

N

=

84.8 g Li

3

N

Lithium produces less lithium nitride than nitrogen gas. Therefore, the limiting reactant is lithium, and the theoretical yield of lithium nitride is

20.9 g

.

Explanation:

6 0
3 years ago
An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t
Dmitry_Shevchenko [17]

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add <u>1.4 L</u> of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

<u>Step 1:</u> Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

<u>Step 2: </u>Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

<u>Step 3:</u> Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

<u>Step 4:</u> The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

<u>Step 5:</u> The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

<u>Step 6: </u>Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.

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