Question

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

Answer 1

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

Step-by-step 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:

• 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:

`Molarity=(number of moles)/(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 H_(2) SO_(4) =(0.005699 moles ofFe(OH)_(3)* 3moles of H_(2) SO_(4) )/(1mole ofFe(OH)_(3))`

moles of H₂SO₄= 0.017097

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

`0.2860M=(0.017097 moles)/(volume)`

Solving for the volume:

`volume=(0.017097 moles)/(0.2860 M)`

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.