# How many moles of sodium hydroxide are needed to react with 148.0 g of Iron (III) bromide?

## Question:

How many moles of sodium hydroxide are needed to react with {eq}148.0 \ g {/eq} of {eq}Iron (III) {/eq} bromide ?

## Reaction Stoichiometry:

Stoichiometry is the relation between the number of moles of reactants and products in a balanced chemical reaction. The coefficients of the reactants and products are directly proportional to their respective number of moles consumed or formed in the reaction.

Given:

Mass of Iron(III) Bromide=148.0 g

The molar mass of Iron(III) Bromide=295.56g/mol

Moles of Iron(III) Bromide=148.0 g/(295.56g/mol)=0.500 mol

The reaction between sodium hydroxide and iron (III) bromide is given by:

{eq}\mathrm{FeBr_{3}+3NaOH \rightarrow FeO(OH)+3NaBr+H_{2}O} {/eq}

Now, stoichiometrically, the molar ratio of NaOH and {eq}\mathrm{FeBr_{3}} {/eq} is 1:3

So, 1 moles of {eq}\mathrm{FeBr_{3}} {/eq} reacts with 3 moles of NaOH.

Thus, 0.500 moles of {eq}\mathrm{FeBr_{3}} {/eq} reacts with {eq}0.5\times 3= 1.5 {/eq} moles of NaOH.