Question

(h) 114.00 g/mol
c₉h₂₀ + o₂ → co₂ + h₂o
(k) 128.00 g/mol

Explanation:

Step1: Balance the chemical equation

The balanced equation for the combustion of $C_9H_{20}$ is $C_9H_{20}+14O_2
ightarrow9CO_2 + 10H_2O$.

Step2: Analyze molar - mass information

The molar mass of $C_9H_{20}$ is given as $128.00\ g/mol$. The other molar - mass value of $114.00\ g/mol$ is not relevant to this reaction as written. But if we were to calculate amounts of reactants and products, we would use the stoichiometry from the balanced equation. For example, if we had a certain mass of $C_9H_{20}$, we could convert it to moles using $n=\frac{m}{M}$ (where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass), and then use the mole - ratios from the balanced equation to find the moles and masses of $O_2$, $CO_2$, and $H_2O$.

Answer:

The balanced chemical equation is $C_9H_{20}+14O_2
ightarrow9CO_2 + 10H_2O$.