What ratio of 73% calcium hypochlorite powder to water mimics the strength of sodium hypochlorite (bleach)? Calcium hypochlorite is sold as pool shock in varying strengths.  I've read that it can be dissolved in water to create a 'bleach' (typically sodium hydrochloric or sodium hydroxide) and is often used as such to treat drinking water in developing countries, but every source I've read prescribes a different ratio. Strictly looking at chlorine levels for water sanitation, cleaning, etc., how much powder (at 73% calcium hydrochlorite) would you need to dissolve into 1 gallon of water to match common household bleach?
To start, we must first understand that we can ignore the counterions to the $\ce{Ca(ClO)2}$ and the $\ce{NaClO}$. Typical household bleach claims to have a concentration of 6% (w/v). Now we can do some basic arithmetic to convert the percent weight by volume of sodium hypochlorite in water to a molar concentration:

$\mathrm{\frac{0.06\,g}{1\,mL}}$$\mathrm{\frac{1\,mol}{74.44\,g}}$$\mathrm{\frac{1000\,mL}{1\,L}}$$\mathrm{\,= 0.8060\,M \,OCl^-}$

The first fraction in the above equation is based on the definition of w/v, which is the number of grams of a substance dissolved in one milliliter of a solution times 100%. In the next fraction, we convert the grams $\ce{NaClO}$ into moles of $\ce{NaClO}$. Finally, in the last fraction, we convert milliliters into liters to give the number of moles per liter of $\ce{NaClO}$, or simply $\ce{ClO^-}$ without the counter ion. 

Now that we know the molar concentration of $\ce{ClO^-}$ we are aiming for, we can convert this into the number of grams of pure $\ce{Ca(ClO)2}$ we would need to to achieve this concentration in one gallon of water

$\mathrm{\frac{0.8060\,mol}{1\,L}}$$\mathrm{\frac{1\,mol}{2\,mol}}$$\mathrm{\frac{142.98\,g}{1\,mol}}$$\mathrm{\frac{3.78541\,L}{1\,gal}}$$\mathrm{\,=\frac{218.1\,g\,OCl^-}{gal}}$

In the second fraction, we converted the moles of $\ce{ClO^-}$ into the number of moles of $\ce{Ca(ClO)2}$ (as there is only one unit of $\ce{Ca(ClO)2}$ per two units of $\ce{ClO^-}$). In the third fraction, we converted the number of moles of $\ce{Ca(ClO)2}$ into grams. In the last fraction, we converted the number of liters into the number of gallons. 

As you stated, the powder only contains 73% $\ce{Ca(ClO)2}$ by mass. We know we need a total of 218.1 g of $\ce{Ca(ClO)2}$, so we can use the following conversion factor:

$\mathrm{\frac{218.1\,g\,OCl^-}{1\,gal}}$$\mathrm{\frac{1\,g}{0.73\,g}}$$\mathrm{\,=298.8\,g\,powder}$ (Rounded to 0.3 kg since we only have one significant figure)

In the second fraction, we know that per every gram of your powder, 73 percent (0.73 g) of the $\ce{Ca(ClO)2}$ will be present. Thus, the conversion factor is valid. 

Final Answer: 0.3 kg of Powder