ron (II) hydroxide, Fe(OH)2, is very insoluble in water. Hence, if it is produced during an aqueous reaction it will precipitate from solution. How many grams of sodium hydroxide (a strong base) must be added to 42.7 mL of a 0.305 M aqueous solution of iron (II) nitrate, in order to precipitate all of the iron as Fe(OH)2

1.04g of NaOH must be added to precipitate all of the iron

Explanation:

Based on the chemical reaction:

Fe(NO₃)₂(aq) + 2NaOH(aq) → Fe(OH)₂(s) + 2NaNO₃(aq)

Where 1 mole of Iron (II) nitrate reacts with 2 moles of NaOH to produce 1 mole of Fe(OH)₂

To solve this question we need to find the moles of Fe(NO₃)₂ and using the chemical reaction we can find the moles of NaOH and its mass:

Moles Fe(NO₃)₂:

0.0427L * (0.305mol / L) = 0.0130moles of Fe(NO₃)₂

Moles NaOH

0.0130moles of Fe(NO₃)₂ * (2mol NaOH / 1mol Fe(NO₃)₂) =

0.0260 moles NaOH

Mass -Molar mass NaOH = 40g/mol-:

0.0260 moles NaOH * (40g / mol) =

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if you mean hydrogen and oxygen then water is the product