Answer:
See explanation
Explanation:
In this question, we have to remember that in all combustion reactions we will have Oxygen as reactive ([tex]O_2[/tex]) and the products are Carbon dioxide ([tex]CO_2[/tex]) and Water ([tex]H_2O[/tex]). Additionally, for the states, we will have (l) for the liquid state and (g) for the gas state. So, we can analyze each reaction:
Part A.
The formula for nonane is [tex]C_9H_2_0[/tex], with this in mind we can write the combustion reaction:
[tex]C_9H_2_0_(_I_)~ +~ O_2~->~CO_2_(_g_)~ +~H_2O_(_g_) [/tex]
When we balance the reaction we will obtain:
[tex]C_9H_2_0_(_I_)~ +~14 O_2_(_g_)~->~9CO_2_(_g_)~ +~10H_2O_(_g_) [/tex]
Part B.
The formula for 2-methylbutane is [tex]C_5H_1_2[/tex], with this in mind we can write the combustion reaction:
[tex]C5H12_(_I_)~ +~O_2_(_g_)~->~CO_2_(_g_)~ +~H2O_(_g_) [/tex]
When we balance the reaction we will obtain:
[tex]C5H12_(_I_)~ +~8O_2_(_g_)~->~5 CO_2_(_g_)~ +~6 H2O_(_g_) [/tex]
Part C.
The formula for 3-ethyltoluene is [tex]C_9H_1_2[/tex], with this in mind we can write the combustion reaction:
[tex]C_9H_1_2_(_I_)~+~O_2_(_g_)~->~CO_2_(_g_)~+~H2O_(_g_)[/tex]
When we balance the reaction we will obtain:
[tex]C9H12_(_I_)~ +~12O_2_(_g_)~->~9CO_2_(_g_)~ +~6H2O_(_g_)[/tex]
See figure 1 for further explanations.
I hope it helps!