Chemistry, 23.07.2019 16:50, Richelleypinelley

Write a balanced equation of the combustion of cooking fat.

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answered: princessmnicole2

Answer : Combustion reactions are those reactions which includes reacting with the oxygen along with the reactant.

Here, the reactant is cooking fat.

$C_{51}H_{88}O_{6} + 70 O_{2} --- 51 CO_{2} + 44 H_{2}O$

So when a long chain hydrocarbon which is cooking fat here reacts with oxygen the products obtained are carbon dioxide and water molecule.

One mole of fat reacts with 70 moles of oxygen and gives 51 moles of carbon dioxide and 44 moles of water molecule.

answered: jwood287375

The balanced equation for the combustion of cooking fat is

$\boxed{{{\text{C}}_{51}}{{\text{H}}_{88}}{{\text{O}}_6}+70{{\text{O}}_2}\to51{\text{C}}{{\text{O}}_{\text{2}}} + 44{{\text{H}}_{\text{2}}}{\text{O}}}$ .

Further explanation:

The chemical reaction that contains an equal number of atoms of the different elements in the reactant as well as in the product side is known as a balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

Combustion reaction:

It is the reaction in which the reactant reacts with molecular oxygen to form carbon dioxide and a water molecule. Molecular oxygen acts as the oxidizing agent in these reactions. A large amount of heat is released and therefore combustion reactions are exothermic in nature.

The steps to balance a chemical reaction are as follows:

Step 1: Complete the reaction and write the unbalanced symbol equation.

In the combustion reaction, ${{\text{C}}_{51}}{{\text{H}}_{88}}{{\text{O}}_6}$ reacts with ${{\text{O}}_2}$ to form ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ . The physical state of ${{\text{C}}_{51}}{{\text{H}}_{88}}{{\text{O}}_6}$ is liquid, ${{\text{O}}_2}$ is gas, ${\text{C}}{{\text{O}}_{\text{2}}}$ is gas and ${{\text{H}}_{\text{2}}}{\text{O}}$ is gas. The unbalanced chemical equation is as follows:

${{\text{C}}_{51}}{{\text{H}}_{88}}{{\text{O}}_6}\left( l \right)+{{\text{O}}_2}\left( g \right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right)+{{\text{H}}_{\text{2}}}{\text{O}}\left( g \right)$

Step2: Then we write the number of atoms of all the different elements that are present in a chemical reaction in the reactant side and product side separately.

⦁On reactant side,

Number of carbon atoms is 51.

Number of hydrogen atoms is 88.

Number of oxygen atoms is 6.

⦁On product side,

Number of carbon atoms is 1.

Number of hydrogen atoms is 2.

Number of oxygen atoms is 3.

Step3: Initially, we try to balance the number of other atoms of elements except for carbon, oxygen, and hydrogen by multiplying with some number on any side but in the combustion reaction there is only carbon, hydrogen, and oxygen atom.

Step 4: After this, we balance the number of atoms of carbon and then hydrogen atom followed by oxygen atoms.

To balance the number of atoms we have to multiply ${{\text{O}}_2}$ by 70, ${\text{C}}{{\text{O}}_{\text{2}}}$ by 51 and ${{\text{H}}_{\text{2}}}{\text{O}}$ by 44. This gives,

${{\text{C}}_{51}}{{\text{H}}_{88}}{{\text{O}}_6}+\boxed{70}{\text{ }}{{\text{O}}_2} \to \boxed{51}{\text{ C}}{{\text{O}}_{\text{2}}}+\boxed{44}{\text{ }}{{\text{H}}_{\text{2}}}{\text{O}}$