Determine the value of X in the hydrated sodium carbonate formula Na2CO3·XH2O given the mass of hydrated and anhydrous forms.

Here's the solution to question 23:

Given:
Mass of hydrated sodium carbonate ($\text{Na}_2\text{CO}_3 \cdot \text{XH}_2\text{O}$) = $2.86 \text{ g}$
Mass of anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$) = $1.06 \text{ g}$

Atomic masses: $\text{Na} = 23.0 \text{ g/mol}$, $\text{C} = 12.0 \text{ g/mol}$, $\text{O} = 16.0 \text{ g/mol}$, $\text{H} = 1.0 \text{ g/mol}$.

Step 1: Calculate the molar mass of anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$).

$$\text{Molar mass of } \text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 46.0 + 12.0 + 48.0 = 106.0 \text{ g/mol}$$

a) The number of moles of anhydrous sodium carbonate in the residue.

$$\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{Mass of } \text{Na}_2\text{CO}_3}{\text{Molar mass of } \text{Na}_2\text{CO}_3} = \frac{1.06 \text{ g}}{106.0 \text{ g/mol}} = 0.0100 \text{ mol}$$

The number of moles of anhydrous sodium carbonate is $\boxed{\text{0.0100 mol}}$.

b) The number of moles of water of crystallization present in the hydrated carbonate.
First, calculate the mass of water lost:

$$\text{Mass of } \text{H}_2\text{O} = \text{Mass of hydrated salt} - \text{Mass of anhydrous salt}$$ $$\text{Mass of } \text{H}_2\text{O} = 2.86 \text{ g} - 1.06 \text{ g} = 1.80 \text{ g}$$

Next, calculate the molar mass of water ($\text{H}_2\text{O}$):

$$\text{Molar mass of } \text{H}_2\text{O} = (2 \times 1.0) + 16.0 = 2.0 + 16.0 = 18.0 \text{ g/mol}$$

Now, calculate the number of moles of water:

$$\text{Moles of } \text{H}_2\text{O} = \frac{\text{Mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{H}_2\text{O}} = \frac{1.80 \text{ g}}{18.0 \text{ g/mol}} = 0.100 \text{ mol}$$

The number of moles of water of crystallization is $\boxed{\text{0.100 mol}}$.

c) Deduce the formula of the hydrated sodium carbonate.
To find the value of X in $\text{Na}_2\text{CO}_3 \cdot \text{XH}_2\text{O}$, we need the mole ratio of $\text{H}_2\text{O}$ to $\text{Na}_2\text{CO}_3$.

$$\text{Ratio } X = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{Na}_2\text{CO}_3} = \frac{0.100 \text{ mol}}{0.0100 \text{ mol}} = 10$$

The formula of the hydrated sodium carbonate is $\boxed{\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}}$.