Draw and name the structure of polymer formed from propene. Write an ionic equation for the reaction between sodium sulphate solution and lead (II) nitrate solution.

15. (b) Draw and name the structure of polymer formed from propene.
Propene is $\text{CH}_2=\text{CH}-\text{CH}_3$. During polymerization, the double bond breaks, and the monomers link together to form polypropene.

The repeating unit of polypropene is:

$$\begin{matrix} & \text{CH}_3 \\ & | \\ - & \text{CH}_2 & - & \text{CH} & - \end{matrix}_n$$

The name of the polymer is $\boxed{\text{Polypropene}}$.

16. (a) Write an ionic equation for the reaction between sodium sulphate solution and lead (II) nitrate solution.
The reaction is between lead (II) nitrate ($\text{Pb(NO}_3)_2$) and sodium sulphate ($\text{Na}_2\text{SO}_4$) to form lead (II) sulphate ($\text{PbSO}_4$) precipitate and sodium nitrate ($\text{NaNO}_3$).

$$\text{Pb(NO}_3)_2 \text{(aq)} + \text{Na}_2\text{SO}_4 \text{(aq)} \longrightarrow \text{PbSO}_4 \text{(s)} + 2\text{NaNO}_3 \text{(aq)}$$

The full ionic equation is:

$$\text{Pb}^{2+} \text{(aq)} + 2\text{NO}_3^- \text{(aq)} + 2\text{Na}^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \longrightarrow \text{PbSO}_4 \text{(s)} + 2\text{Na}^+ \text{(aq)} + 2\text{NO}_3^- \text{(aq)}$$

The net ionic equation, removing spectator ions, is:

$$\boxed{\text{Pb}^{2+} \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \longrightarrow \text{PbSO}_4 \text{(s)}}$$

(b) Determine the mass of the lead salt formed in the reaction in (a) above.
The lead salt formed is lead (II) sulphate ($\text{PbSO}_4$). The initial lead powder (0.63 g) is converted entirely into $\text{Pb}^{2+}$ ions, which then form $\text{PbSO}_4$.