Question

what is the coefficient in front of the o₂ when the equation below is balanced? c₂h₆ (g) + o₂ (g) → co₂ (g) + h₂o (g) use only whole numbers and the lowest whole - number ratio.

Explanation:

Step1: Balance carbon atoms

There are 2 carbon atoms in $C_2H_6$. So, put 2 in front of $CO_2$: $C_2H_6(g)+O_2(g)\to2CO_2(g) + H_2O(g)$.

Step2: Balance hydrogen atoms

There are 6 hydrogen atoms in $C_2H_6$. So, put 3 in front of $H_2O$: $C_2H_6(g)+O_2(g)\to2CO_2(g)+3H_2O(g)$.

Step3: Balance oxygen atoms

On the right - hand side, there are $2\times2 + 3\times1=7$ oxygen atoms. So, put $\frac{7}{2}$ in front of $O_2$: $C_2H_6(g)+\frac{7}{2}O_2(g)\to2CO_2(g)+3H_2O(g)$.

Step4: Get whole - number coefficients

Multiply all coefficients by 2 to get whole - number coefficients: $2C_2H_6(g)+7O_2(g)\to4CO_2(g)+6H_2O(g)$.

Answer:

7