Draw the dot and cross structures for NH3, H3O+, and NH4OH.

a) Ammonia molecule ($NH_3$)
Nitrogen has 5 valence electrons, and each hydrogen has 1 valence electron. Nitrogen forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons.

$$\begin{array}{c} \text{H} \\ \text{.} \\ \text{H} \cdot \times \text{N} \times \cdot \text{H} \\ \text{.} \\ \text{.} \end{array}$$

(Where 'x' represents electrons from Nitrogen and '.' represents electrons from Hydrogen. The two dots above N represent the lone pair.)

b) Hydronium ion ($H_3O^+$)
Oxygen has 6 valence electrons, and each hydrogen has 1 valence electron. The +1 charge means one electron is lost. Oxygen forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons.

$$\begin{bmatrix} \begin{array}{c} \text{H} \\ \text{.} \\ \text{H} \cdot \times \text{O} \times \cdot \text{H} \\ \text{.} \\ \text{.} \end{array} \end{bmatrix}^+$$

(Where 'x' represents electrons from Oxygen and '.' represents electrons from Hydrogen. The two dots above O represent the lone pair.)

c) Ammonium hydroxide ($NH_4OH$)
This is an ionic compound consisting of the ammonium ion ($NH_4^+$) and the hydroxide ion ($OH^-$).

Ammonium ion ($NH_4^+$)
Nitrogen has 5 valence electrons, and each hydrogen has 1 valence electron. The +1 charge means one electron is lost. Nitrogen forms four single covalent bonds with four hydrogen atoms and has no lone pairs.

$$\begin{bmatrix} \begin{array}{c} \text{H} \\ \text{.} \\ \text{H} \cdot \times \text{N} \times \cdot \text{H} \\ \text{.} \\ \text{H} \end{array} \end{bmatrix}^+$$

(Where 'x' represents electrons from Nitrogen and '.' represents electrons from Hydrogen.)

Hydroxide ion ($OH^-$)
Oxygen has 6 valence electrons, and hydrogen has 1 valence electron. The -1 charge means one electron is gained. Oxygen forms one single covalent bond with hydrogen and has three lone pairs of electrons.

$$\begin{bmatrix} \text{.} \\ \text{.} \\ \text{H} \cdot \times \text{O} \times \\ \text{.} \\ \text{.} \end{bmatrix}^-$$

(Where 'x' represents electrons from Oxygen and '.' represents electrons from Hydrogen. The six dots around O represent the three lone pairs.)

The reaction is between $XCO_3$ and hydrochloric acid ($HCl$).
The balanced chemical equation is:

$$XCO_3(s) + 2HCl(aq) \rightarrow XCl_2(aq) + H_2O(l) + CO_2(g)$$

Calculate the moles of $HCl$:
Volume of $HCl = 20 \text{ cm}^3 = 0.020 \text{ L}$
Concentration of $HCl = 1.8 \text{ M}$

$$\text{Moles of } HCl = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of } HCl = 1.8 \text{ mol/L} \times 0.020 \text{ L} = 0.036 \text{ mol}$$

From the balanced equation, the mole ratio of $XCO_3$ to $HCl$ is $1:2$.

$$\text{Moles of } XCO_3 = \frac{\text{Moles of } HCl}{2}$$ $$\text{Moles of } XCO_3 = \frac{0.036 \text{ mol}}{2} = 0.018 \text{ mol}$$

Calculate the molar mass of $XCO_3$:
Mass of $XCO_3 = 1.312 \text{ g}$

$$\text{Molar mass of } XCO_3 = \frac{\text{Mass}}{\text{Moles}}$$ $$\text{Molar mass of } XCO_3 = \frac{1.312 \text{ g}}{0.018 \text{ mol}} \approx 72.89 \text{ g/mol}$$

Calculate the R.A.M of X:
Molar mass of $XCO_3 = \text{R.A.M of X} + \text{R.A.M of C} + (3 \times \text{R.A.M of O})$
Given C = 12.0, O = 16.0.

$$72.89 = \text{R.A.M of X} + 12.0 + (3 \times 16.0)$$ $$72.89 = \text{R.A.M of X} + 12.0 + 48.0$$ $$72.89 = \text{R.A.M of X} + 60.0$$ $$\text{R.A.M of X} = 72.89 - 60.0$$ $$\text{R.A.M of X} = 12.89$$

The R.A.M of X is approximately 12.9.

a) Potassium iodide and lead nitrate

$$Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)$$

b) Magnesium sulphate and sodium carbonate

$$Mg^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MgCO_3(s)$$

c) Calcium carbonate and dilute nitric (V) acid

$$CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g)$$