Question

Question:A sample of liquid heptane (C₇H₁₆) weighing 11.50 g is reacted with 1.300 moles of oxygen gas.The heptane is burned completely (heptane reacts with oxygen to form both carbon monoxideand water and carbon dioxide and water). After the reaction is complete, the amount of gaspresent is 1.050 moles (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?

Short answer

1. The moles of CO produced are 0.42 moles.
2. The moles of CO₂producedare 0.385 moles.
3. There are 0.245 moles of O₂are left.

Step-by-step solution

Balanced equation for burning of heptane

The balanced equation can be denoted as:

$$\begin{gathered} C_7{H}_{16\left(1\right)}+11O_{2\left(g\right)}\to 7C{O}_{2\left(g\right)}+8H_2{O}_{\left(1\right)} \\ 2C_7{H}_{16\left(1\right)}+15O_{2\left(g\right)}\to 14C{O}_{\left(g\right)}+16H_2{O}_{\left(1\right)} \end{gathered}$$

Here, Heptane’s mass is 11.5 g. On converting to the moles number,

$n=\frac{m}{M}$

The mole number is n; m is mass and molar mass is M.

$$\begin{gathered} M(C_{7}H{}_{16})=7.Ar(C)+16Ar(H) \\ =7\times 12+1\times 16 \\ =100\frac{g}{mol} \\ n(C_7{H}_{16})=\frac{11.5g}{100\frac{g}{mol}} \\ n(C_7{H}_{16})=0.115 mol \end{gathered}$$

Number of gaseous product formation using mole ratio

The ratio of C₇H₁₆ (l) to gas product is 1:7. The gases are CO and CO₂. The gaseous product number can be found as,

$0.115 mol heptane.\frac{7 mol gaseous products}{1 mol heptane}=0.805 mol CO+C{O}_2$

The amount of gas is 1.05 mole. Left over can be found as

$1.05 mol-0.805 mol=0.245 mol of O_2$

Using system of equation

To find mole number, first find $n\left(O_2\right)$ which reacts with heptane.

$1.3 mol O_2.0.245 mol O_2=1.055 mol O_2 reacts$

Using these equations, 11 mol of oxygen reacts with 1 mole of C₇H₁₆ for CO₂ production. There are two moles of C₇H₁₆ (l) reacts with 15 moles of oxygen gas for CO production.

$$\begin{gathered} \left(C_7{H}_{16}\right(1\left)\right):n\left(O_2\right)=1:11 \\ n\left(O_2\right)=11n\left(C_7{H}_{16}\right(1\left)\right) \\ n\left(C_7{H}_{16}\right(2\left)\right):n\left(O_2\right)=2:15 \\ n\left(O_2\right)=7.5n\left(C_7{H}_{16}\right(2\left)\right) \end{gathered}$$

Total number of moles can be found as 0.115.

$$\begin{gathered} \left(C_7{H}_{16}\right(1\left)\right)+n\left(C_7{H}_{16}\right(2\left)\right)=0.115 \\ 11n\left(C_7{H}_{16}\right(1\left)\right)+7.5n\left(C_7{H}_{16}\right(2\left)\right)=1.055 \end{gathered}$$

Finding the number of moles for CO2 and CO

Solve the equations to find the number of moles produced:

$$\begin{gathered} n_{C_7{H}_{16}{\left(l\right)}_1}=0.055 mol \\ {n}_{C_7{H}_{16}{\left(l\right)}_2}=0.06 mol \\ {n}_{CO}=0.06\times 7 \\ {n}_{CO}=0.42 mol \\ {n}_{C{O}_2}=0.055\times 7 \\ {n}_{C{O}_2}=0.385 mol \end{gathered}$$

The mass of CO produced are 0.42 moles, and moles of CO₂produced are 0.385 moles. The moles of O₂ left are 0.245 moles.