Question

1. when the equation for the complete combustion of propene (c₃h₆(g)) is balanced using the lowest whole numbers, the coefficient on oxygen is

□ 4
□ 9
□ 6
□ 2

Explanation:

Step1: Write the un - balanced combustion equation

The combustion of propene ($C_3H_6$) in oxygen ($O_2$) produces carbon dioxide ($CO_2$) and water ($H_2O$). The un - balanced equation is $C_3H_6(g)+O_2(g)
ightarrow CO_2(g)+H_2O(g)$.

Step2: Balance carbon atoms

There are 3 carbon atoms in $C_3H_6$. So, we put a coefficient of 3 in front of $CO_2$: $C_3H_6(g)+O_2(g)
ightarrow3CO_2(g)+H_2O(g)$.

Step3: Balance hydrogen atoms

There are 6 hydrogen atoms in $C_3H_6$. So, we put a coefficient of 3 in front of $H_2O$: $C_3H_6(g)+O_2(g)
ightarrow3CO_2(g)+3H_2O(g)$.

Step4: Balance oxygen atoms

On the right - hand side, the total number of oxygen atoms is $3\times2 + 3\times1=9$. So, we put a coefficient of $\frac{9}{2}$ in front of $O_2$: $C_3H_6(g)+\frac{9}{2}O_2(g)
ightarrow3CO_2(g)+3H_2O(g)$.

Step5: Multiply through by 2 to get whole numbers

Multiply each coefficient by 2 to get the balanced equation with the lowest whole numbers: $2C_3H_6(g)+9O_2(g)
ightarrow6CO_2(g)+6H_2O(g)$.

Answer:

9