${\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)$. The ratio of moles of CO₂ and H₂O in the above balanced equation is :

The unbalanced chemical equation is C₄​H_10​(g)+O_2​(g)→CO2​(g)+H2​O(l)

On LHS, we have 4C atoms. So on RHS we add coefficient of 4 to CO₂​ to balance C atoms.
 C₄​H_10​(g)+O_2​(g)→4CO_2​(g)+H₂​O_(l)

On LHS, we have 10H atoms. To balance H atoms, we add coefficient 5 to H₂​O
 C₄​H_10​(g)+O_2​(g)→4CO_2​(g)+5H₂​O_(l)

On LHS, we have 2 O atoms. On RHS, we have 13 O atoms. To balance O atoms, we add $\frac{13}{2}$ to O₂​ on LHS
 C₄​H_10​(g)+$\frac{13}{2}$​O_2​(g)→4CO_2​(g)+5H_2​O_(l)

This is balanced chemical equation. For  using the lowest whole-number terms, we multiply this equation with 2.
 2C₄​H_10​(g)+13O_2​(g)→8CO_2​(g)+10H₂​O_(l)