Describe how a sample of Iron (II) Carbonate can be prepared using dilute hydrochloric acid.

Question 4(a): Number of moles of HCl used

Step 1: Formula for number of moles
$n = c \times V$
where $n$ is number of moles (mol), $c$ is concentration (mol dm$^{-3}$), $V$ is volume (dm$^3$).

Step 2: Convert volume to dm$^3$
$V = \frac{25}{1000} = 0.025\ \textrm{dm}^3$

Step 3: Substitute values
$n(\ce{HCl}) = 0.1\ \textrm{mol dm}^{-3} \times 0.025\ \textrm{dm}^3 = 0.0025\ \textrm{mol}$

0.0025 mol

Question 4(b): Number of moles of NaOH that reacted with HCl

Step 1: Moles of HCl used in back titration (equals moles of excess NaOH, since 1:1 reaction)
$n(\ce{HCl, back}) = 0.1\ \textrm{mol dm}^{-3} \times \frac{24.9}{1000}\ \textrm{dm}^3$
$n(\ce{HCl, back}) = 0.1 \times 0.0249 = 0.00249\ \textrm{mol}$
Excess $\ce{NaOH} = 0.00249\ \textrm{mol}$.

Step 2: Total moles of NaOH added
$n(\ce{NaOH, total}) = n(\ce{HCl, pipetted}) + n(\ce{HCl, back})$
$n(\ce{NaOH, total}) = 0.0025\ \textrm{mol} + 0.00249\ \textrm{mol} = 0.00499\ \textrm{mol}$

Step 3: Moles of NaOH that reacted with HCl
$n(\ce{NaOH, reacted}) = n(\ce{NaOH, total}) - \textrm{excess NaOH}$
$n(\ce{NaOH, reacted}) = 0.00499\ \textrm{mol} - 0.00249\ \textrm{mol} = 0.0025\ \textrm{mol}$

0.0025 mol

Question 4(c): Concentration of NaOH in moles per dm$^3$

Step 1: Volume of NaOH added in dm$^3$
$V(\ce{NaOH}) = \frac{25.0}{1000} = 0.0250\ \textrm{dm}^3$

Step 2: Concentration formula
$c = \frac{n}{V}$

Step 3: Substitute values
$c(\ce{NaOH}) = \frac{0.00499\ \textrm{mol}}{0.0250\ \textrm{dm}^3} = 0.20\ \textrm{mol dm}^{-3}$

0.20 mol dm^{-3}