Q4.138CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: In 1997, at the United Nations Conference on Climate Change, the major industrial nations agreed to expand their research efforts to develop renewable sources of carbon-based fuels. For more than a decade, Brazil has been engaged in a program to replace gasoline with ethanol derived from the root crop manioc (cassava).

(a) Write separate balanced equations for the complete combustion of ethanol (C₂H₅OH) and of gasoline (represented by the formula C₈H₁₈).

(b) What mass of oxygen is required to burn completely 1.00 L of a mixture that is 90.0% gasoline (d = 0.742 g/mL) and 10.0% ethanol (d = 0.789 g/mL) by volume?

(c) If 1.00 mol of O₂ occupies 22.4 L, what volume of O₂ is needed to burn 1.00 L of the mixture?

(d) Air is 20.9% O₂ by volume. What volume of air is needed to burn 1.00 L of the mixture?

Step-by-Step Solution

Answer

 

Answer:

1. The balanced equation for the complete combustion of ethanol and gasoline are,  $$\begin{gathered} C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right) \\ 2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)\to 16C{O}_2\left(g\right)+18H_{2}O\left(l\right) \end{gathered}$$
2. The mass of the oxygen gas is 2.51x10³ g.
3. The volume of O₂ is 1.76x10³ L.
4. The volume of the air is 8.42x10³ L.

 

Step 1: (a) Reactions

The balanced equation for the complete combustion of ethanol is,

 $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

 The balanced equation for the combustion of gasoline is,

 $2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)\to 16C{O}_2\left(g\right)+18H_{2}O\left(l\right)$

Step 2: (b) Determination of mass of oxygen

The equation for the complete combustion of ethanol is,

 $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

 

The volume of ethanol is,

 $$\begin{gathered} V=1.00L\times \frac{10}{100}\times \frac{1000mL}{1L} \\ V=100.0mL \end{gathered}$$

 

Now, the mass of oxygen gas using the volume of ethanol is,

 $$\begin{gathered} =\left(100.0mL\right)\times \frac{0.789g C_2{H}_{5}OH}{1mL C_2{H}_{5}OH}\times \frac{1mol C_2{H}_{5}OH}{46.0g C_2{H}_{5}OH}\times \frac{3mol O_2}{1mol C_2{H}_{5}OH}\times \frac{32.0g O_2}{1mol O_2}\backslash \mathrm{hfill} \\ =165.0g \end{gathered}$$

 

 The equation for the combustion of gasoline is,

 $2C_8{H}_{18}\left(l\right)+25O_2\left(g\right)\to 16C{O}_2\left(g\right)+18H_{2}O\left(l\right)$

 

The volume of gasoline is,

 $$\begin{gathered} V=1.00L\times \frac{90}{100}\times \frac{1000mL}{1L} \\ V=900.0mL \end{gathered}$$

 

Now, the mass of oxygen gas using the volume of ethanol is,

$$\begin{gathered} =\left(900.0mL\right)\times \frac{0.742g C_8{H}_{18}}{1mL C_8{H}_{18}}\times \frac{1mol C_8{H}_{18}}{114.0g C_8{H}_{18}}\times \frac{12.5mol O_2}{1mol C_8{H}_{18}}\times \frac{32.0g O_2}{1mol O_2}\backslash \mathrm{hfill} \\ =2.34\times {10}^{3}g \end{gathered}$$ 

 

Now, the total mass of the oxygen gas is,

 $$\begin{gathered} 165.0g+\left(2.34\times {10}^{3}g\right) \\ =2.51\times {10}^{3}g \end{gathered}$$

 

Thus, the mass of the oxygen gas is 2.51x10³ g.