( Use ↑ and ↓ please ! )
Butane $C_4H_{10}$ reacts with oxygen $O_2$ to form carbon dioxide $CO_2$ and water $H_2O$ . Equation?
Reagents: $..C_4H_{10}$ $+$ $..O_2$ $\longrightarrow$ $...$
Products: $..C_4H_{10}$ $+$ $..O_2$ $\longrightarrow$ $..CO_2$ $+$ $..H_2O$
Let's take one molecule butane: $\color{red}{1}C_4H_{10}$ $+$ $..O_2$ $\longrightarrow$ $..CO_2$ $+$ $..H_2O$
One molecule $C_3H_8$ contains 4 atoms $C$, one molecule $CO_2$ only one. 4 molecules $CO_2$ will be produced: $1C_4H_{10}$ $+$ $..O_2$ $\longrightarrow$ $\color{red}{4}CO_2$ $+$ $..H_2O$
One molecule $C_4H_{10}$ contains 10 atoms $H$, one molecule $H_2O$ two. 5 molecules $H_2O$ will be produced: $1C_3H_8$ $+$ $..O_2$ $\longrightarrow$ $4 CO_2$ $+$ $\color{red}{5}H_2O$
4 molecules $CO_2$ and 5 molecules $H_2O$ contain 13 atoms $O$, one molecule $O_2$ two. $\frac{13}{2}$ molecules $O_2$ would be necessary! Half-molecules do not exist.Let's multiply all by 2: $2C_4H_{10}$ $+$ $\color{red}{13}O_2$ $\longrightarrow$ $8 CO_2$ $+$ $10 H_2O$
Balanced equation : $2C_4H_{10}$ $+$ $13O_2$ $\longrightarrow$ $8 CO_2$ $+$ $10 H_2O$