Write an equation for the combustion of methanol (CH3OH) and calculate the number of moles of methanol used.

Here are the answers to your questions:

a) Write an equation for the combustion of methanol ($\text{CH}_3\text{OH}$).

Step 1: Write the unbalanced equation for complete combustion, which produces carbon dioxide and water.

$$\text{CH}_3\text{OH(l)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$$

Step 2: Balance the carbon atoms. (1 C on each side, already balanced).

Step 3: Balance the hydrogen atoms. There are 4 H atoms on the left (3 from $\text{CH}_3$ and 1 from $\text{OH}$). So, place a coefficient of 2 in front of $\text{H}_2\text{O}$ on the right.

$$\text{CH}_3\text{OH(l)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$$

Step 4: Balance the oxygen atoms. There are $1 + 2x$ oxygen atoms on the left (1 from $\text{CH}_3\text{OH}$ and $2x$ from $\text{O}_2$). There are $2 + (2 \times 1) = 4$ oxygen atoms on the right.

$$1 + 2x = 4 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}$$

Step 5: Substitute the coefficient for $\text{O}_2$ and multiply the entire equation by 2 to remove the fraction.

$$2\text{CH}_3\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$$

The balanced equation is:

$$\boxed{2\text{CH}_3\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}}$$

b) Calculate:

i) The number of moles of methanol used in this experiment.

Step 1: Calculate the mass of methanol used.

$$\text{Mass of methanol used} = \text{Initial mass of lamp + methanol} - \text{Final mass of lamp + methanol}$$ $$\text{Mass of methanol used} = 22.98\text{ g} - 22.11\text{ g} = 0.87\text{ g}$$

Step 2: Calculate the molar mass of methanol ($\text{CH}_3\text{OH}$).

$$\text{Molar mass of CH}_3\text{OH} = (1 \times \text{C}) + (4 \times \text{H}) + (1 \times \text{O})$$ $$$$