how many grams of pure iron can be recovered from 50 grams of an ore which is known to be 32% ferrous nitrate?

how many grams of pure iron can be recovered from 50 grams of an ore which is known to be 32% ferrous nitrate? Again I think I need to do this in two ways. I‘m thinking Double the grams to 100 or do I just subtract 32% from 100? It‘s been a while since I have done this

50g ore x 0.32 16g Fe(NO3)2 16g Fe(NO3)2 / 179.84g/mole 0.089moles Fe(NO3)2 Fe(NO3)2 -- Fe + NO2 + 2O2 0.089moles Fe(NO3)2 yields 0.089moles Fe 0.089moles Fe x 55.84g/mole 4.97g

Nausea and constipation.

multiply 50 times .32 to obtain amount of ferrous nitrate: 50*.3216grams iron(ii)nitrate the formula wt is approx 179.8gm per mol, atomic wt of iron is 55.8gm so: nitrogen 14gm oxygen16 gm 55.8/179.8 * 16 4.96 grams of pure iron

mass Fe(NO3)2 50 x 32 /10016 g moles Fe(NO3)2 16 g /179.86 g/mol0.089 moles Fe2+ 0.089 mass Fe2+ 0.089 mol x 55.847 g/mol4.97 g ( 5.0 g at two significant figures) Note : ferrous nitrate is Fe(NO3)2 ferric nitrate is Fe(NO3)3

constipation and green stools