Propane (C3H8) + Oxygen (O2) = CO2 + H2O  How to balance?
To balance a chemical equation, the number of atoms of each element must be equal on both sides(reactant and product sides) of the equation.
Here is a stepbystep process to balance the equation “C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O”:
Steps to balance the equation “C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O”
Step 1: Write the unbalanced equation: ⇒ C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O 
Step 2: Count the number of atoms of each element on both sides of the equation. Look at the number of atoms of each element on the reactant side (left side) and compare it to the number of atoms of that element on the product side (right side).

Step 3: Balance the Carbon atoms In the above equation, we can see that we have 3 Carbon atoms on the reactant side and 1 Carbon atom on the product side. So, to balance the number of Carbon atoms, we need to multiply the number of CO_{2} on the product side by 3. The new equation will be: ⇒ C_{3}H_{8} + O_{2} → 3CO_{2} + H_{2}O Again compare the number of atoms on the reactant side to the product side.
[∴ Carbon atoms balanced in the above equation] 
Step 4: Balance the Hydrogen atoms Similarly, according to the above equation(step 3), we have 8 Hydrogen atoms on the reactant side and 2 Hydrogen atoms on the product side. So, to balance the number of Hydrogen atoms, we need to multiply the number of H_{2}O on the product side by 4. Now the equation looks like this: ⇒ C_{3}H_{8} + O_{2} → 3CO_{2} + 4H_{2}O Again compare the number of atoms on the reactant side to the product side.
[∴ Hydrogen atom balanced in this equation] 
Step 5: Balance the Oxygen atoms Now we have to balance the Oxygen atoms, according to the Step 4 equation, we have 2 Oxygen atoms on the reactant side and (6 + 4 = 10) Oxygen atoms on the product side. So, we need to multiply the oxygen atoms on the reactant side by 5 to balance the number of Oxygen atoms. The final balanced equation will be: ∴ C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O Again compare the number of atoms on the reactant side to the product side.
[∴ All the atoms balanced in this equation] 
So, the balance equation of the reaction: C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O is – ⇒ C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O [Answer]

In this stepbystep solution, we have balanced the equation: (C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O) by adjusting the coefficient of each element, ensuring that the number of atoms of each element is the same on both sides of the equation.
FAQ
What type of reaction is C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O? 
The reaction of C_{3}H_{8} (propane) and O_{2} (oxygen) to form CO_{2} (carbon dioxide) and H_{2}O (water) is a combustion reaction. A combustion reaction is a type of chemical reaction in which a fuel (in this case, C_{3}H_{8} or propane) reacts with an oxidizer (O_{2} or oxygen) to produce heat and light (in the form of a flame) as well as new products (CO_{2} and H_{2}O). The balanced chemical equation for this reaction (C_{3}H_{8} + O_{2}) is C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O. The overall equation represents the complete chemical change of the reactants into the products. In this reaction, C_{3}H_{8} is the fuel, and O_{2} is the oxidizer. The heat and light produced in this reaction are due to the breaking and forming of chemical bonds between the atoms in the reactants and products. 
How many grams of O_{2} are needed to completely burn 33 g of C_{3}H_{8}? 
The balanced chemical equation for this reaction (C_{3}H_{8} + O_{2}) is – ⇒ C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O. This equation tells us that for every 1 mole of C_{3}H_{8} that is burned, 5 moles of O_{2} are consumed. To determine the number of grams of O_{2} needed to completely burn 33 grams of C_{3}H_{8}, we need to convert the mass of C_{3}H_{8} to moles and then multiply by the stoichiometric coefficient of O_{2}. First, we can convert the mass of C_{3}H_{8} to moles using its molar mass, which is 44.1 g/mol. ⇒ Number of moles of C_{3}H_{8} = Mass/molar mass = 33 g/44.1 g/mol = 0.747 mol C_{3}H_{8} Then, we can multiply this value by the stoichiometric coefficient of O_{2} from the balanced equation to find the number of moles of O_{2} required. ⇒ 0.747 * 5 = 3.735 mol O_{2} Now, we can convert the number of moles of O_{2} back to grams using its molar mass, which is 32.0 g/mol. ⇒ 3.735 * 32.0 g/mol = 119.52 g O_{2} So, approximately 119.52 g of O_{2} is needed to completely burn 33 g of C_{3}H_{8}. 
What coefficient would the O_{2} have after balancing C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O? 
As we already know, the balancing equation for the reaction (C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O) is – ⇒ C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O. According to the above equation, the coefficient of O_{2} would be 5. This means that for every 1 mole of C_{3}H_{8} that is burned, 5 moles of O_{2} are consumed. 
How many moles of oxygen (O_{2}) is necessary to react with 4 moles of propane (C_{3}H_{8})? 
First, let’s start with the balanced chemical equation for the combustion of propane: ⇒ C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O. This equation tells us that for every 1 mole of propane (C_{3}H_{8}) that is burned, 5 moles of O_{2} are consumed. We are given the 4 moles of Propane (C_{3}H_{8}) and we know that for every 1 mole of propane that is burned, 5 moles of O_{2} are consumed. So, to react with 4 moles of propane (C_{3}H_{8}), we would need – ∴ 4 moles of propane * 5 moles of O_{2}/1 mole of propane = 20 moles of O_{2}. So, the 20 moles of oxygen (O_{2}) necessary to react with 4 moles of propane (C_{3}H_{8}). 
Here check the balancing equation (C_{3}H_{8} + O_{2} → CO_{2} + H_{2}O) video to clear your doubt –
References
 ChemicalAid. “Equation Balancer.” [Online]. Available: https://www.chemicalaid.com/tools/equationbalancer.php?equation=C3H8+%2B+O2+%3D+CO2+%2B+H2O&hl=en. [Accessed: 25Jan2023].
 Socratic. “Balance the equation: C3H8 + O2 = CO2 + H2O.” [Online]. Available: https://socratic.org/questions/balancetheequationc3h8o2co2h2o. [Accessed: 25Jan2023].
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