Propane (C3H8) + Oxygen (O2) = CO2 + H2O - How to balance?

Step 1: Write the unbalanced equation:

⇒ C₃H₈ + O₂ → CO₂ + H₂O

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 CO₂ on the product side by 3.

The new equation will be:

⇒ C₃H₈ + O₂ → 3CO₂ + H₂O

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 H₂O on the product side by 4.

Now the equation looks like this:

⇒ C₃H₈ + O₂ → 3CO₂ + 4H₂O      

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:

∴ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O    

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: C₃H₈ + O₂ → CO₂ + H₂O is –

⇒ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O    [Answer]

What type of reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O?

The reaction of C₃H₈ (propane) and O₂ (oxygen) to form CO₂ (carbon dioxide) and H₂O (water) is a combustion reaction.

A combustion reaction is a type of chemical reaction in which a fuel (in this case, C₃H₈ or propane) reacts with an oxidizer (O₂ or oxygen) to produce heat and light (in the form of a flame) as well as new products (CO₂ and H₂O).

The balanced chemical equation for this reaction (C₃H₈ + O₂) is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

The overall equation represents the complete chemical change of the reactants into the products. In this reaction, C₃H₈ is the fuel, and O₂ is the oxidizer.

How many grams of O₂ are needed to completely burn 33 g of C₃H₈?

The balanced chemical equation for this reaction (C₃H₈ + O₂) is –

⇒ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

This equation tells us that for every 1 mole of C₃H₈ that is burned, 5 moles of O₂ are consumed.

To determine the number of grams of O₂ needed to completely burn 33 grams of C₃H₈, we need to convert the mass of C₃H₈ to moles and then multiply by the stoichiometric coefficient of O₂.

First, we can convert the mass of C₃H₈ to moles using its molar mass, which is 44.1 g/mol.

⇒ Number of moles of C₃H₈ = Mass/molar mass = 33 g/44.1 g/mol = 0.747 mol C₃H₈

Then, we can multiply this value by the stoichiometric coefficient of O₂ from the balanced equation to find the number of moles of O₂ required.

⇒ 0.747 * 5  = 3.735 mol O₂

Now, we can convert the number of moles of O₂ back to grams using its molar mass, which is 32.0 g/mol.

⇒ 3.735 * 32.0 g/mol = 119.52 g O₂

So, approximately 119.52 g of O₂ is needed to completely burn 33 g of C₃H₈.

What coefficient would the O₂ have after balancing C₃H₈ + O₂ → CO₂ + H₂O?

As we already know, the balancing equation for the reaction (C₃H₈ + O₂ → CO₂ + H₂O) is –

⇒ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

According to the above equation, the coefficient of O₂ would be 5. This means that for every 1 mole of C₃H₈ that is burned, 5 moles of O₂ are consumed.

How many moles of oxygen (O₂) is necessary to react with 4 moles of propane (C₃H₈)?

First, let’s start with the balanced chemical equation for the combustion of propane:

⇒ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

This equation tells us that for every 1 mole of propane (C₃H₈) that is burned, 5 moles of O₂ are consumed.

We are given the 4 moles of Propane (C₃H₈) and we know that for every 1 mole of propane that is burned, 5 moles of O₂ are consumed.

So, to react with 4 moles of propane (C₃H₈), we would need –

∴ 4 moles of propane * 5 moles of O₂/1 mole of propane

= 20 moles of O₂.

So, the 20 moles of oxygen (O₂) necessary to react with 4 moles of propane (C₃H₈).