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What is electrolysis? Write the structural formula, balanced equation for complete combustion and molar mass of butane.

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2 parts
a

58 g/mol\text{58 g/mol}

0.2 mol\text{0.2 mol}

c

Cu2+, SO42−, H+, OH−\text{Cu}^{2+}\text{, SO}_4^{2-}\text{, H}^{+}\text{, OH}^{-}

12000 C\text{12000 C}

0.124 F\text{0.124 F}

Step-by-step solution

8 steps

Part a· 4 steps

Butane (C₄H₁₀) is the main component of cooking gas.

  • i) Structural formula of butane:

    HHHH∣∣∣∣H−C−C−C−C−H∣∣∣∣HHHH\begin{matrix} \text{H} & \text{H} & \text{H} & \text{H} \\ | & | & | & | \\ \text{H} - \text{C} & - \text{C} & - \text{C} & - \text{C} - \text{H} \\ | & | & | & | \\ \text{H} & \text{H} & \text{H} & \text{H} \end{matrix}

  • ii) Balanced equation for complete combustion of butane:

    2C4H10(g)+13O2(g)⟶8CO2(g)+10H2O(g)\text{2C}_4\text{H}_{10}\text{(g)} + \text{13O}_2\text{(g)} \longrightarrow \text{8CO}_2\text{(g)} + \text{10H}_2\text{O(g)}

  • iii) Calculate the molar mass of butane (C=12, H=1):

  1. Identify the number of atoms for each element.

    Butane has 4 carbon atoms and 10 hydrogen atoms.

  2. Calculate the molar mass.

    $$
\text{Molar mass of C}_4\text{H}_{10} = (4 \times 12) + (10 \times 1)
$$
$$
= 48 + 10
$$
$$
= \boxed{\text{58 g/mol}}
$$
    • iv) How many moles are present in 11.6 g of Butane?

  3. Use the formula for moles.

    $$
\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}
$$

  4. Substitute the given values.

    $$
\text{Moles} = \frac{11.6 \text{ g}}{58 \text{ g/mol}}
$$
$$
= \boxed{\text{0.2 mol}}
$$

Part b

**What is electrolysis?

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction. It involves passing an electric current through an electrolyte (a molten or dissolved ionic compound) to decompose it into its constituent elements or other compounds.

Part c

Cu2+, SO42−, H+, OH−\text{Cu}^{2+}\text{, SO}_4^{2-}\text{, H}^{+}\text{, OH}^{-}

A copper (II) sulphate solution was electrolyzed using inert electrodes.

  • i) State the different ions present in the solution: The ions present are from copper (II) sulphate and water.

    Cu2+, SO42−, H+, OH−\boxed{\text{Cu}^{2+}\text{, SO}_4^{2-}\text{, H}^{+}\text{, OH}^{-}}

  • ii) Given that a steady current of 2 A was passed through the solution for 1 hour 40 minutes, calculate the quantity of charge passed:

  1. Convert time to seconds.

    $$
\text{Time (t)} = 1 \text{ hour } 40 \text{ minutes}
$$
$$
= (1 \times 60 \text{ minutes}) + 40 \text{ minutes}
$$
$$
= 60 \text{ minutes} + 40 \text{ minutes} = 100 \text{ minutes}
$$
$$
= 100 \times 60 \text{ seconds} = 6000 \text{ s}
$$

  2. Calculate the quantity of charge (Q) using the formula Q=I×tQ = I \times t.

    $$
Q = 2 \text{ A} \times 6000 \text{ s}
$$
$$
Q = \boxed{\text{12000 C}}
$$
    • iii) The number of Faradays used:

  3. Use Faraday's constant (1 F=96485 C1 \text{ F} = 96485 \text{ C}).

    $$
\text{Number of Faradays} = \frac{\text{Total charge (Q)}}{\text{Faraday's constant (F)}}
$$

  4. Substitute the calculated charge.

    $$
\text{Number of Faradays} = \frac{12000 \text{ C}}{96485 \text{ C/mol}}
$$
$$
\approx \boxed{\text{0.124 F}}
$$
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