Question

question 9
select the substances and coefficients to correctly complete and balance
the equation for the complete combustion of propene, c₃h₆.
c₃h₆(g) + → +

Explanation:

Step1: Write the general combustion equation

For a hydrocarbon \(C_xH_y\), the general combustion equation is \(C_xH_y + O_2
ightarrow CO_2 + H_2O\). For propene \(C_3H_6\), the initial 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 \(CO_2\). The equation becomes \(C_3H_6(g)+O_2(g)
ightarrow 3CO_2(g)+H_2O(g)\).

Step3: Balance hydrogen atoms

There are 6 hydrogen atoms in \(C_3H_6\). So, we put a coefficient of 3 in front of \(H_2O\). The equation is \(C_3H_6(g)+O_2(g)
ightarrow 3CO_2(g)+3H_2O(g)\).

Step4: Balance oxygen atoms

On the right - hand side, there are \(3\times2 + 3\times1=9\) oxygen atoms. So, we put a coefficient of \(\frac{9}{2}\) in front of \(O_2\). But we usually prefer whole - number coefficients. Multiply the entire equation by 2: \(2C_3H_6(g)+9O_2(g)
ightarrow 6CO_2(g)+6H_2O(g)\).

Answer:

\(2C_3H_6(g)+9O_2(g)
ightarrow 6CO_2(g)+6H_2O(g)\)