Making a 10L HCl solution with 8.77% HCl in a 100 mL solution from a 16 M stock solution
I figured out the problem I had. I needed a way to convert grams to milliliters, which was density.

I found the density of HCl for the 8.77% concentration:
$$ HCl \ Density = 1.04 g/mL$$
I then found how much HCl would be in 100g of solution:
$$ 8.77\% \ of \ 100g, \ so \ 8.77g \ HCl \ and \ 91.23g \ solvent$$
I did some dimensional analysis to get to moles divided by liters, so that I could get the molarity:
$$ (\frac{8.77 g \ HCl}{1})*(\frac{1 \ mol}{36.461g})*(\frac{1}{100g \ solution})*(\frac{1.04g}{1 mL})*(\frac{1000 mL}{1 L}) = 2.5 M$$
Then, I did a nice dilution calculation:
$$ 10 L *2.5M = 16M *L_2$$
$$ \frac{25}{16} = 1.5625 L$$

To do the dilution I would place 1.5625 L HCl into a beaker, and then fill with solvent until I reached a total solution volume of 10 L.