How do you balance the equation: C4div0 + O2 → CO2 + H2O

Balancing a chemical equation involves making sure that the same number of each type of atom appears on both sides of the equation. Let's balance the equation for the combustion of butane, which is represented as:

C₄H₁₀ + O₂ → CO₂ + H₂O

Here's the step-by-step process to balance this equation:

Count the number of atoms of each element on both sides of the equation:

Carbon (C): 4 on the left, 1 on the right (in CO₂)
Hydrogen (H): 10 on the left, 2 on the right (in H₂O)
Oxygen (O): 2 on the left (in O₂), 2 on the right (in CO₂) and 1 on the right (in H₂O).
Start by balancing the carbon atoms. We need 4 CO₂ molecules on the right side to match the 4 carbon atoms on the left side:
C₄H₁₀ + O₂ → 4CO₂ + H₂O

Next, balance the hydrogen atoms. We have 10 hydrogen atoms on the left and only 2 on the right, so we need to add a coefficient of 5 in front of H₂O on the right side to balance the hydrogen atoms:
C₄H₁₀ + O₂ → 4CO₂ + 5H₂O

Now, let's balance the oxygen atoms. On the left side, we have 2 oxygen atoms in O₂, and on the right side, we have 4 carbon atoms in CO₂ (which contributes 4 oxygen atoms) and 10 hydrogen atoms in 5H₂O (which contributes 5 oxygen atoms, as there are 2 oxygen atoms in each water molecule). In total, we have 4 + 5 = 9 oxygen atoms on the right side. To balance this, we need to adjust the coefficient in front of O₂ on the left side. Since we already have 2 oxygen atoms in O₂, we need to add 7 more O₂ molecules to the left side:
C₄H₁₀ + 7O₂ → 4CO₂ + 5H₂O

Now, the equation is balanced with respect to the number of each type of atom on both sides. There are 4 carbon atoms, 10 hydrogen atoms, and 18 oxygen atoms on both sides of the equation.