Question

a sample of liquid heptane (c₇h₁₆) weighing 11.10 g is reacted with 1.330 mol of oxygen gas. the heptane is burned (heptane reacts with oxygen to form both carbon monoxide and water and carbon dioxide). after the reaction is complete, no heptane remains and the amount of gas present is 1.040 mol (assume that all of the water formed is liquid). (a) how many moles of co are produced? (b) how many moles of co₂ are produced? (c) how many moles of o₂ are left over?

Explanation:

Step1: Write the balanced - chemical equation

The combustion of heptane ($C_7H_{16}$) is $C_7H_{16}(l)+11O_2(g)
ightarrow7CO_2(g) + 8H_2O(l)$.

Step2: Calculate the moles of heptane

The molar mass of $C_7H_{16}$ is $M=(7\times12.01 + 16\times1.01)\ g/mol=100.23\ g/mol$. The moles of $C_7H_{16}$, $n_{C_7H_{16}}=\frac{m}{M}=\frac{11.10\ g}{100.23\ g/mol}=0.1107\ mol$.

Step3: Determine the moles of products and reactants based on stoichiometry

(a) For CO

From the balanced equation, the mole - ratio of $C_7H_{16}$ to $CO$ is 1:7. But we assume complete combustion to $CO_2$ first. Since there is no information about incomplete combustion to $CO$, we assume $n_{CO}=0\ mol$.

(b) For $CO_2$

The mole - ratio of $C_7H_{16}$ to $CO_2$ is 1:7. So, $n_{CO_2}=0.1107\ mol\times7 = 0.7749\ mol$.

(c) For $O_2$ left over

The mole - ratio of $C_7H_{16}$ to $O_2$ is 1:11. The moles of $O_2$ required for complete combustion of $0.1107\ mol$ of $C_7H_{16}$ is $n_{O_2\ required}=0.1107\ mol\times11 = 1.2177\ mol$. Given $n_{O_2\ initial}=1.330\ mol$. The moles of $O_2$ left over, $n_{O_2\ left}=1.330\ mol - 1.2177\ mol=0.1123\ mol$.

Answer:

(a) 0 mol
(b) 0.7749 mol
(c) 0.1123 mol