AlCl_{3}, named Aluminium chloride, is a Lewis acid, and H_{2}SO_{4} (sulfuric acid) is a strong inorganic acid. Let us examine their reaction in detail.

**H _{2}SO_{4}, also known as the “oil of vitriol”, is a powerful dehydrating agent used in various organic syntheses. AlCl_{3} is highly hygroscopic and forms hydrates when reacts with moisture.**

In this article, we will discuss the reaction products, enthalpy, and molecular forces involved in the reaction of H_{2}SO_{4} and AlCl_{3}.

## What is the product of H_{2}SO_{4} and AlCl_{3}

**Al _{2}(SO_{4})_{3} (Aluminium sulfate) and HCl (hydrochloric acid) are the products of the reaction H_{2}SO_{4} + AlCl_{3}.**

**H _{2}SO_{4} + AlCl_{3 }→ Al_{2}(SO_{4})_{3 }+ HCl**

## What type of reaction is H_{2}SO_{4} + AlCl_{3}

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

## How to balance H_{2}SO_{4} + AlCl_{3}

**The balanced equation is derived using the below steps**

**H _{2}SO_{4} + AlCl_{3 }→ Al_{2}(SO_{4})_{3 }+ HCl**

**The elements present in the reactants and the products are counted.**

Elements involved | Reactants side | Products side |
---|---|---|

H | 2 | 1 |

Al | 1 | 2 |

S | 1 | 3 |

O | 4 | 12 |

Cl | 3 | 1 |

**Number of elements**

**In the next step, coefficients are used to balance the number of elements on each side of the equation. A coefficient of 2, 3, and 6 are added before AlCl**_{3}, H_{2}SO_{4,}and HCl, respectively.

**Thus, the balanced equation is**

**3H**_{2}SO_{4}+ 2AlCl_{3 }→ Al_{2}(SO_{4})_{3 }+ 6HCl

## H_{2}SO_{4} + AlCl_{3} titration

**H _{2}SO_{4} + AlCl_{3 }titration is not possible as H_{2}SO_{4} and AlCl_{3} both are acidic in nature so H_{2}SO_{4} cannot be used to titrate against AlCl_{3}.**

## H_{2}SO_{4} + AlCl_{3} net ionic equation

**The net ionic equation for H _{2}SO_{4} + AlCl_{3} is**

**6H ^{+}(aq)+ 3SO_{4}^{2-}(aq) + 2Al^{3+}(aq) + 6Cl^{–}(aq)→ 2Al^{3+}(aq) + 3SO_{4}^{2-}(aq)+ 6H^{+}(aq) + 6Cl^{–}(aq)**

** The net ionic equation is obtained using the following steps**

**The balanced equation is written in the first step.****3H**_{2}SO_{4}+ 2AlCl_{3 }→ Al_{2}(SO_{4})_{3 }+ 6HCl**The phases (solid, liquid, gas, or aqueous) is indicated for the species involved in the reaction.****3H**_{2}SO_{4}(aq) + 2AlCl_{3}(aq)→ Al_{2}(SO_{4})_{3}(aq)+ 6HCl(aq)**The strong electrolytes are split into their corresponding ions.****6H**^{+}(aq)+ 3SO_{4}^{2-}(aq) + 2Al^{3+}(aq) + 6Cl^{–}(aq)→ 2Al^{3+}(aq) + 3SO_{4}^{2-}(aq)+ 6H^{+}(aq) + 6Cl^{–}(aq)**In the above equation all the ions are balanced so the net ionic equation is****6H**^{+}(aq)+ 3SO_{4}^{2-}(aq) + 2Al^{3+}(aq) + 6Cl^{–}(aq)→ 2Al^{3+}(aq) + 3SO_{4}^{2-}(aq)+ 6H^{+}(aq) + 6Cl^{–}(aq)

## H_{2}SO_{4} + AlCl_{3} conjugate pairs

**H _{2}SO_{4} + AlCl_{3} will be a conjugate pair as**

**H _{2}SO_{4} + Cl^{–} = SO_{4}^{2-} + HCl**

**SO**_{4}^{2-}is the conjugate base of acid H_{2}SO_{4}.**HCl is formed as the conjugate acid of Cl**^{–}**(base)**.

## H_{2}SO_{4} and AlCl_{3} intermolecular forces

**The intermolecular forces found in the H _{2}SO_{4} and AlCl_{3 }are,**

**Dispersion forces, hydrogen bonding, and dipole-dipole interactions exist between the molecules of H**_{2}SO_{4}.**Electrostatic interactions are present in AlCl**_{3}molecules as the compound is ionic in nature.

## H_{2}SO_{4} + AlCl_{3} reaction enthalpy

**H _{2}SO_{4} + AlCl_{3 }reaction enthalpy is -140.54 KJ/mol. The enthalpy is calculated using the values listed below and by putting them in the formula,**

Compounds involved | Enthalpy in KJ/mol |
---|---|

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

HCl | -92.3 |

AlCl_{3} | -705.63 |

Al_{2}(SO_{4})_{3} | -3440 |

**Values of enthalpy**

**∆H**_{f}^{°}(reaction) = ∆H_{f}^{°}(products) – ∆H_{f}^{°}(reactants)

**= -3993.8 – (-3853.26)**

**= -140.54 KJ/mol**

## Is H_{2}SO_{4} + AlCl_{3} a buffer solution

**H _{2}SO_{4} + AlCl_{3} will not make a buffer solution as the acid H_{2}SO_{4} is a strong acid whereas weak acid is required to make a buffer.**

## Is H_{2}SO_{4} + AlCl_{3} a complete reaction

**H _{2}SO_{4} + AlCl_{3} is a complete reaction as the reactants are completely consumed to form the products and the products are not undergoing any further reaction.**

## Is H_{2}SO_{4} + AlCl_{3} an exothermic or endothermic reaction

**H _{2}SO_{4} + AlCl_{3 }is an exothermic reaction due to the liberation of heat during the reaction**

**and the enthalpy is also negative for the reaction.**

## Is H_{2}SO_{4} + AlCl_{3} a redox reaction

**H _{2}SO_{4} + AlCl_{3 }is not a redox reaction as the oxidation states of the elements do not change after the reaction.**

## Is H_{2}SO_{4} + AlCl_{3} a precipitation reaction

**H _{2}SO_{4} + AlCl_{3} is not a precipitation reaction as the products obtained are completely soluble in water and no precipitate is formed.**

## Is H_{2}SO_{4} + AlCl_{3} reversible or irreversible reaction

**H _{2}SO_{4} + AlCl_{3} is a reversible reaction because HCl is more acidic than H_{2}SO_{4} so the backward reaction is more feasible as the equilibrium shift towards a less acidic product**.

## Is H_{2}SO_{4} + AlCl_{3} displacement reaction

**H _{2}SO_{4} + AlCl_{3} is a double-displacement reaction where the aluminium being more reactive displaces the hydrogen from its salt to form Al_{2}(SO_{4})_{3} and the H^{+} ion combines with chloride ion to give HCl.**

**H _{2}SO_{4} + AlCl_{3 }= Al_{2}(SO_{4})_{3 }+ HCl**

#### Conclusion

The reaction is exothermic and reversible. Al_{2}(SO_{4})_{3} produced is widely used as a mordant in the dyeing industry and has high importance in the purification of water and removal of phosphorus from wastewater. HCl is a strong acid and has immense application in synthetic chemistry.