To calculate the volume of the 29.2% (w/w) HCl stock solution needed to prepare a 200 mL solution of 0.4 M HCl, we need to use several steps involving concentration and density conversions.

First, we calculate the mass of HCl that is contained in the 200 mL of a 0.4 M solution:

$$ Mass = Molarity \times Volume \times Molecular\ Weight $$

$$ Mass = 0.4 \ mol/L \times 0.200 \ L \times 36.5 \ g/mol $$

Note that we convert the volume from mL to L to match the units of molarity (mol/L).

Now, we calculate it:

$$ Mass = (0.4 \times 0.200 \times 36.5) \ g $$

$$ Mass = 0.08 \times 36.5 \ g $$

$$ Mass = 2.92 \ g $$

The next step is to determine how much of the stock solution is needed to get 2.92 g of HCl. Since the stock solution is 29.2% (w/w) HCl, this means that in every 100 g of stock solution, there is 29.2 g of HCl. We can set up a proportion to find the mass of the stock solution needed:

$$ \frac{29.2\ g \ HCl}{100\ g \ stock\ solution} = \frac{2.92\ g \ HCl}{x\ g \ stock\ solution} $$

Now we solve for $$ x $$:

$$ x = \frac{2.92\ g \times 100\ g \ stock\ solution}{29.2\ g \ HCl} $$

$$ x = \frac{292}{29.2} \ g $$

$$ x = 10\ g $$

So, we need 10 g of the stock solution to get 2.92 g of HCl.

The final step is to calculate the volume of the stock solution that has a mass of 10 g. We use the density to convert mass to volume:

$$ Volume = \frac{Mass}{Density} $$

The density of the stock solution is given as 1.25 g/mL, so:

$$ Volume = \frac{10\ g}{1.25\ g/mL} $$

$$ Volume = 8\ mL $$

Therefore, to prepare a 200 mL solution of 0.4 M HCl, you would need to measure out 8 mL of the 29.2% HCl stock solution.