H_{2}SO_{4} (Sulphuric acid) is a strong acid and Fe_{3}O_{4} (Iron oxide) occurs in nature as the mineral magnetite. Let us know some facts about the H_{2}SO_{4} + Fe_{3}O_{4}.

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**Sulphuric acid is also known as oil of vitriol. It is colorless, odorless viscous liquid. Iron (II, III) oxide is composed with the elements Oxygen and Iron. It is one of the oxides of the iron.**

In this article we will discuss about the reaction between Sulphuric acid and Iron (II, III) oxide, balancing of equation and net ionic equation of the reaction.

**What is the product of H**_{2}SO_{4} and Fe_{3}O_{4}

_{2}SO

_{4}and Fe

_{3}O

_{4}

**Ferric sulphate [Fe _{2}(SO_{4})_{3}], Ferrous sulphate (FeSO_{4}) and Water (H_{2}O) are produced in the reaction between H_{2}SO_{4} and Fe_{3}O_{4}**.

**4H _{2}SO_{4} + Fe_{3}O_{4} = Fe_{2}(SO_{4})_{3} + FeSO_{4} + 4H_{2}O**

**What type of reaction is H**_{2}SO_{4} + Fe_{3}O_{4}

_{2}SO

_{4}+ Fe

_{3}O

_{4}

**H _{2}SO_{4} + Fe_{3}O_{4} is a double displacement reaction**.

**How to tát balance H**_{2}SO_{4} + Fe_{3}O_{4}

_{2}SO

_{4}+ Fe

_{3}O

_{4}

**The reaction is balanced using following steps**:

**H _{2}SO_{4} + Fe_{3}O_{4} = Fe_{2}(SO_{4})_{3} + FeSO_{4} + H_{2}O**

**The number of atoms present on both sides is counted which should be the same.**

Reactant Side | Product Side |
---|---|

2-Hydrogen atom | 2-Hydrogen atom |

8-Oxygen atom | 17-Oxygen atom |

1-Sulphur atom | 4-Sulphur atom |

3-Iron atom | 3-Iron atom |

**Table representing number of atoms present in reactant side and product side**

**The numbers of Sulpur atoms of both the sides of the reaction balance by adding coefficient 4 to tát the Sulphuric acid molecule at reactant side**.**4H**_{2}SO_{4}+ Fe_{3}O_{4}= Fe_{2}(SO_{4})_{3}+ FeSO_{4}+ H_{2}O**The number of Oxygen atoms of both the sides of the reaction balance by adding coefficient 4 to tát the water molecule at product side.****4H**_{2}SO_{4}+ Fe_{3}O_{4}= Fe_{2}(SO_{4})_{3}+ FeSO_{4}+ 4H_{2}O**So the balanced chemical reaction is:****4H**_{2}SO_{4}+ Fe_{3}O_{4}= Fe_{2}(SO_{4})_{3}+ FeSO_{4}+ 4H_{2}O

**H**_{2}SO_{4} + Fe_{3}O_{4 }Titration

_{2}SO

_{4}+ Fe

_{3}O

_{4 }Titration

**H _{2}SO_{4} + Fe_{3}O_{4} tự not undergo titration because there is no equivalence point**

**H**_{2}SO_{4} + Fe_{3}O_{4} net ionic **equation**

_{2}SO

_{4}+ Fe

_{3}O

_{4}net ionic

**The net ionic equation** **is**

**Fe _{3}O_{4}(s) = 2Fe^{3+}(aq) + Fe^{2+} + 4O^{2-}**

**This net ionic equation is derived using the steps mentioned below**

**The balanced equation with phases for H**_{2}SO_{4}+ Fe_{3}O_{4}**4H**_{2}SO_{4}(aq) + Fe_{3}O_{4}(s) = Fe_{2}(SO_{4})_{3}(aq) + FeSO_{4}(aq) + 4H_{2}O(aq)**Then the equation written in splitting of compounds****4H**_{2}^{+}(aq) + 4SO_{4}^{2-}(aq) + Fe_{3}O_{4}(s) = 2Fe^{3+}(aq) + 3SO_{4}^{2-}(aq) + Fe^{2+}(aq) + SO_{4}^{2-}(aq) + 4H_{2}^{+}(aq) + 4O^{2-}(aq)**The net ionic equation is as follows by removing the spectator ions (ions those are same on reactant & product sides).****Fe**_{3}O_{4}(s) = 2Fe^{3+}(aq) + Fe^{2+}(aq) + 4O^{2-}(aq)

**H**_{2}SO_{4} + Fe_{3}O_{4 }conjugate pairs

_{2}SO

_{4}+ Fe

_{3}O

_{4 }conjugate pairs

**The conjugate pairs of H _{2}SO_{4} + Fe_{3}O_{4} are as follows:**

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**The conjugate base of H**_{2}SO_{4}is SO_{4}^{2-}

**The conjugate pair of Fe**_{3}O_{4}is O_{4}^{2-}

**H**_{2}SO_{4} and Fe_{3}O_{4} Intermolecular forces

_{2}SO

_{4}and Fe

_{3}O

_{4}Intermolecular forces

**Intermolecular forces present in H _{2}SO_{4} + Fe_{3}O_{4} are as follows:**

**Hydogen bonding, dipole-dipole interaction and Van der Waals dispersion are the intermolecular forces in H**_{2}SO_{4}.**Iron (II, III) oxide is ionicaly bonded to tát each other.**

**H**_{2}SO_{4} + Fe_{3}O_{4} reaction enthalpy

_{2}SO

_{4}+ Fe

_{3}O

_{4}reaction enthalpy

**The reaction enthalpy of H _{2}SO_{4} + Fe_{3}O_{4} is -282.97**

**KJ/mol.**

Compound | Moles | Enthalpy of formation, ΔH^{0}_{f} (KJ/mol) |
---|---|---|

H_{2}SO_{4} | 4 | -814 |

Fe_{3}O_{4} | 1 | -1118.4 |

Fe_{2}(SO_{4})_{3} | 1 | -2585.2 |

FeSO_{4} | 1 | -928.85 |

H_{2}O | 4 | -285.83 |

**Bond enthalpy values**

**The standard enthalpy of a reaction is calculated using the formula:**

**ΔH**^{0}_{f (reaction)}= ΣΔH^{0}_{f (product)}– ΣΔH^{0}_{f (reactants)}

**Thus, Enthalpy change =****[1*(-2585.2) + 1*(-928.85) + 4*(-285.83)] – [4*(-814.0) + 1*(-1118.4)]****KJ/mol****= -282.97****KJ/mol**

**Is** **H**_{2}SO_{4} + Fe_{3}O_{4} a buffer solution

_{2}SO

_{4}+ Fe

_{3}O

_{4}a buffer solution

**H _{2}SO_{4} + Fe_{3}O_{4} reaction does not sườn a buffer solution because Sulphuric acid is a strong acid and there can only be present a weak acid or base in buffer solution.**

**Is** **H**_{2}SO_{4} + Fe_{3}O_{4} a complete reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}a complete reaction

**H _{2}SO_{4} + Fe_{3}O_{4 }is a complete reaction and the products lượt thích Ferric sulphate, Ferrous sulphate and water are formed after completion of the reaction.**

**Is** **H**_{2}SO_{4} + Fe_{3}O_{4} an exothermic or endothermic reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}an exothermic or endothermic reaction

**H _{2}SO_{4} + Fe_{3}O_{4 }is an exothermic reaction because calculated enthalpy is negative**

**that is**

**-282.97**

**KJ/mol.**

**Is** **H**_{2}SO_{4} + Fe_{3}O_{4} a redox reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}a redox reaction

**H _{2}SO_{4} + Fe_{3}O_{4 }is a redox reaction due to tát change in oxidation state of iron (Fe) ion on reactant and product side.**

**4H _{2}^{+}+ 4SO_{4}^{2-} + Fe_{3}^{8/3}O_{4} = 2Fe^{3+} + 3SO_{4}^{2-} + Fe^{2+} + SO_{4}^{2-} + 4H_{2}^{+ }+ 4O^{2-}**

**Is H**_{2}SO_{4} + Fe_{3}O_{4} a precipitation reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}a precipitation reaction

**H _{2}SO_{4} + Fe_{3}O_{4 }reaction is not a precipitation reaction because no precipitate is formed during the reaction.**

**Is H**_{2}SO_{4} + Fe_{3}O_{4} a reversible or irreversible reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}a reversible or irreversible reaction

**H _{2}SO_{4} + Fe_{3}O_{4 }is an irreversible reaction because the product formed tự not combine to tát reform reactants under the same condition.**

**Is H**_{2}SO_{4} + Fe_{3}O_{4} a displacement reaction

_{2}SO

_{4}+ Fe

_{3}O

_{4}a displacement reaction

**H _{2}SO_{4} + Fe_{3}O_{4} is a double displacement reaction, in which sulphate ion is displaced from H_{2}SO_{4} molecule and oxygen ion from Fe_{3}O_{4} molecule to tát sườn Ferric sulphate and water.**

**Conclusion**

Sulphuric acid reacts with Ferric oxide it gives Ferric sulphate, Ferrous sulphate and Water. It is a double displacement as well as redox reaction, it founds negative reaction enthalpy hence it is exothermic reaction.

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