How many mL of 0.2860 M sulfuric acid are required to react with 58.42 mL of 0.09756 M iron(III) hydroxide

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

Fe(OH)₃: 1 mole
H₂SO₄: 3 moles
Fe(HSO₄)₃: 1 mole
H₂O: 3 moles

Being molarity (M) the number of moles of solute that are dissolved in a given volume:

[tex]Molarity=\frac{number of moles}{volume}[/tex]

then the number of moles can be calculated as:

number of moles= molarity* volume

So, if 58.42 mL ( 0.05842 L) of 0.09756 M iron (III) hydroxide react, then it means that the number of moles that react are:

number of moles= 0.09756 M* 0.05842 L

number of moles= 0.005699 moles

Now it is possible to apply the following rule of three: if by stoichiometry 1 mole of Fe(OH) ₃ reacts with 3 moles of H₂SO₄, then 0.005699 moles of Fe(OH) ₃ with how many moles of H₂SO₄ does it react?

[tex]moles of H_{2} SO_{4} =\frac{0.005699 moles ofFe(OH)_{3}* 3moles of H_{2} SO_{4} }{1mole ofFe(OH)_{3}}[/tex]

moles of H₂SO₄= 0.017097

Being sulfuric acid 0.2860 M, it is replaced in the definition of molarity:

[tex]0.2860M=\frac{0.017097 moles}{volume}[/tex]

Solving for the volume:

[tex]volume=\frac{0.017097 moles}{0.2860 M}[/tex]

volume= 0.05978 mL

0.05978 mL of 0.2860 M sulfuric acid is required to react with 58.42 mL of 0.09756 M iron (III) hydroxide.