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 and water). 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 a 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 $C_7H_{16}+11O_2
ightarrow 7CO_2 + 8H_2O$. First, find 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 heptane, $n_{C_7H_{16}}=\frac{m}{M}=\frac{11.10\ g}{100.23\ g/mol}=0.1107\ mol$.

Step2: Determine the limiting reactant

From the balanced equation, the mole - ratio of $C_7H_{16}$ to $O_2$ is $1:11$. The moles of $O_2$ required to react completely with $0.1107\ mol$ of $C_7H_{16}$ is $n_{O_2\ required}=0.1107\ mol\times11 = 1.2177\ mol$. Given $n_{O_2\ initial}=1.330\ mol$. So, $C_7H_{16}$ is the limiting reactant.

Step3: Calculate the moles of $CO$ (assuming some incomplete combustion)

There is no information about incomplete combustion in the problem statement and the balanced equation is for complete combustion. In complete combustion of heptane, no $CO$ is produced. So, the moles of $CO$ produced is $0\ mol$.

Step4: Calculate the moles of $CO_2$

From the balanced equation, the mole - ratio of $C_7H_{16}$ to $CO_2$ is $1:7$. Since $C_7H_{16}$ is the limiting reactant, the moles of $CO_2$ produced, $n_{CO_2}=0.1107\ mol\times7 = 0.7749\ mol$.

Step5: Calculate the moles of $O_2$ left

The moles of $O_2$ that reacted is $n_{O_2\ reacted}=0.1107\ mol\times11 = 1.2177\ mol$. The moles of $O_2$ left, $n_{O_2\ left}=n_{O_2\ initial}-n_{O_2\ reacted}=1.330\ mol - 1.2177\ mol=0.1123\ mol$.

Answer:

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