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yan [13]
3 years ago
8

What mass of sulphuric acid can be made from 8g of SO3

Chemistry
1 answer:
Serhud [2]3 years ago
8 0

Answer:

10 g H₂SO₄

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

Given:   8 g SO₃

RxN:   SO₃ + H₂O → H₂SO₄

<u>Step 2: Identify Conversions</u>

Molar Mass of S - 32.07 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of SO₃ - 32.07 + 3(16.00) = 80.07 g/mol

Molar mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol

<u>Step 3: Stoichiometry</u>

<u />8 \ g \ SO_3(\frac{1 \ mol \ SO_3}{80.07 \ g \ SO_3} )(\frac{1 \ mol \ H_2SO_4}{1 \ mol \ SO_3} )(\frac{98.09 \ g \ H_2SO_4}{1 \ mol \ H_2SO_4} ) = 9.80042 g H₂SO₄

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

9.80042 g H₂SO₄ ≈ 10 g H₂SO₄

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Explanation:

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As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

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Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

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Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

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                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

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4 years ago
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