An elixir of ferrous sulfate contains 220 mg of ferrous sulfate in each 5 ml. if each milligram of ferrous sulfate contains the equivalent of 0.2 mg of elemental iron, how many milligrams of elemental iron would be represented in each 5 ml of the elixir?

To determine the mass of elemental iron in the elixir, we need to understand the given values given above. We are given the following:

Mass of ferrous sulfate = 220 mg
Volume of the elixir = 5 mL
Mass of elemental iron / mass of ferrous sulfate = 0.2 mg Fe / mg ferrous sulfate

Therefore, from the given values, we simply multiply the mass of the ferrous sulfate in the elixir and the equivalent mass of elemental iron per mass of ferrous sulfate. We do as follows:

Mass of Fe = 220 mg ferrous sulfate  ( 0.2 mg Fe / mg ferrous sulfate )
Mass of Fe = 44 mg

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o