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sergejj [24]
3 years ago
11

Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the

equation:
N2H4(l) + O2(g) → N2(g) + 2H2O(g)

At STP, if 4.20L of O2 reacts with N2H4, how many liters of water vapor will be produced?

A)
2.10L H2O(g)


B)
2.67L H2O(g)


C)
5.33L H2O(g)


D)
8.40L H2O(g)
Chemistry
1 answer:
lilavasa [31]3 years ago
3 0

Answer:

D)  8.40 L H₂O(g).

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>2N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), </em>

It is clear that 2.0 moles of N₂H₄ react with 1.0 mole of O₂ to produce 2.0 moles of N₂ and 2.0 moles of H₂O.

  • At STP, 4.20L of O₂ reacts with N₂H₄:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u>using cross multiplication: </u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 4.2 L.

∴ 4.2 L of O₂ represents = (1.0 mol)(4.2 L)/(22.4 L) = 0.1875 mol.

  • To find the no. of moles of H₂O produced:

Using cross multiplication:

1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

∴ The no. of moles of H₂O = (2.0 mol)(0.1875 mol)/(1.0 mol) = 3.75 mol.

  • Again, using cross multiplication:

1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

<em></em>

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(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is  1.3 x 10^{-5} .

The initial concentration of propanoic acid is  .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

 

The expression for acid dissociation constant is,

                                                            …… (1)

Here,

 is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

 is the equilibrium concentration of propanoic acid.

ICE table (1):

 

Refer ICE table (1),

 

Substitute the values form the ICE table (1) in equation (1).

 

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

 

Rearrange above equation for x.

                                                                                                           …… (2)

Substitute   for   in equation (2) to calculate the value of x.

 

Therefore, from the ICE table (1) the concentration of hydronium ion is,

 .

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

                                                                                                               …… (3)

Substitute    for    in equation (3) to calculate the pH of the solution.

 

(b)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate  is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

                                                        …… (4)

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

                                                       …… (5)

Dissociation reaction for water is written as follows:

                                                                                       …… (6)

From equation (4), (5), and (6) the relationship between   and   is,

                                                                                                                              …… (7)

Substitute   for   and   for   in equation (7).

 

ICE table (2):

 

The expression for base dissociation constant is,

                                                                                                     …… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

 is the equilibrium concentration of propanoic acid.

From the ICE table (2),

 

Substitute the values form the ICE table (2) in equation (8).

 

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

 

Rearrange above equation for y.

                                                                                                           …… (9)

Substitute   for   in equation (9) to calculate the value of y.

 

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

 

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

                                                                                                           …… (10)

Substitute    for    in equation (10) to calculate pOH of the solution.

 

The relation between pH and pOH is as follows:

                                                                                                                   …… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

 

(c)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

 

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

                                                                                               …… (12)

The negative logarithm of acid ionization constant is equal to  .

                                                                                                                …… (13)

Substitute   for  in equation (13).

 

Substitute    for  ,   for   and 4.9 for    in equation (12).

 

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

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