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

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

Explanation:

The balanced reaction between tha surfuric acid and iron (III) hydroxide is:

Fe(OH)₃ + 3 H₂SO₄ → Fe(HSO₄)₃ + 3 H₂O

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:

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

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?

moles of H₂SO₄= 0.017097

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

Solving for the volume:

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.

58750.08068 is the answer i think