Write a balanced equation for the complete combustion of butane and calculate its molar mass.

Let's solve question 4 parts (a)(ii), (a)(iii), (a)(iv), and (c)(ii)(I), (c)(ii)(II).

Question 4(a)(ii): Write a balanced equation to show the complete combustion of butane.

Step 1: Write the unbalanced equation for the complete combustion of butane ($\text{C}_{4}\text{H}_{10}$).

$$\text{C}_{4}\text{H}_{10} + \text{O}_{2} \longrightarrow \text{CO}_{2} + \text{H}_{2}\text{O}$$

Step 2: Balance the carbon atoms.

$$\text{C}_{4}\text{H}_{10} + \text{O}_{2} \longrightarrow 4\text{CO}_{2} + \text{H}_{2}\text{O}$$

Step 3: Balance the hydrogen atoms.

$$\text{C}_{4}\text{H}_{10} + \text{O}_{2} \longrightarrow 4\text{CO}_{2} + 5\text{H}_{2}\text{O}$$

Step 4: Balance the oxygen atoms.

$$\text{C}_{4}\text{H}_{10} + \frac{13}{2}\text{O}_{2} \longrightarrow 4\text{CO}_{2} + 5\text{H}_{2}\text{O}$$

Step 5: Multiply the entire equation by 2 to remove the fraction.

$$\boxed{2\text{C}_{4}\text{H}_{10} + 13\text{O}_{2} \longrightarrow 8\text{CO}_{2} + 10\text{H}_{2}\text{O}}$$

Question 4(a)(iii): Calculate the molar mass of butane given that C=12 and H=1.

Step 1: Identify the atomic masses and the chemical formula of butane ($\text{C}_{4}\text{H}_{10}$).
Atomic mass of C = $12 \text{ g/mol}$
Atomic mass of H = $1 \text{ g/mol}$

Step 2: Calculate the molar mass.

$$\text{Molar mass of C}_{4}\text{H}_{10} = (4 \times 12 \text{ g/mol}) + (10 \times 1 \text{ g/mol})$$ $$\text{Molar mass of C}_{4}\text{H}_{10} = 48 \text{ g/mol} + 10 \text{ g/mol}$$ $$\text{Molar mass of C}_{4}\text{H}_{10} = 58 \text{ g/mol}$$

The molar mass of butane is:

$$\boxed{58 \text{ g/mol}}$$

Question 4(a)(iv): How many moles are present in 11.6 g of Butane?

Step 1: Use the formula for the number of moles.

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