Question

7.110 Gasohol is a fuel containing liquid ethanol (C₂H₂O) thatburns in oxygen gas to give carbon dioxide and water gases.(7.4, 7.7,7.8)a. Write the balanced chemical equation.b. How many moles of O₂ are needed to completely react with 4.0 moles of C₂H₂O?c. If a car produces 88 g of CO2, how many grams of O₂ are used up in the reaction?d. If you burn 125 g of C₂H₂O, how many grams of CO₂ and H₂O can be produced?

Answer 1

a. C₂H₂O + 2O₂ → 2CO₂ + H₂O

b. 8 moles of O₂

c. 64 g of O₂

d. The grams of CO₂ produced is 262 g and the mass of H₂O produced is 53.6 g.

Step-by-step explanation

a. The balanced chemical equation gasohol is a fuel containing liquid ethanol (C₂H₂O) that burns in oxygen gas to give carbon dioxide and water gases is:

`C_2H_2O+2O_2\rightarrow2CO_2+H_2O`

b. The moles of O₂ needed to completely react with 4.0 moles of C₂H₂O?.

The mole ratio of O₂ to C₂H₂O in the balanced equation is 2:1.

Therefore, 4.0 moles of C₂H₂O will completely react with (2 x 4) = 8 moles of O₂.

c. The mass in grams of O₂ that is used up in the reaction if a car produces 88 g of CO₂.

The molar mass of CO₂ = 44.01 g/mol

The molar mass of O₂ = 31.999 g/mol

The mole ratio of O₂ to CO₂ is 2:2

This implies (2 mol x 31.999 g/mol) = 63.998 g of O₂ is used up to produce (2 mol x 44.01 g/mol) = 88.02 g of CO₂

Therefore x g of O₂ will be used up to produce 88 g of CO₂

That is;

`\begin{gathered} 63.998g\text{ }O₂=88.02g\text{ }CO₂ \\ \\ x=88g\text{ }CO₂ \\ \\ Cross\text{ }multiply\text{ }and\text{ }divide\text{ }both\text{ }sides\text{ }by\text{ }88.02g\text{ }CO₂ \\ \\ x=\frac{88g\text{ }CO₂}{88.02\text{ }CO₂}*63.998g\text{ }O₂ \\ \\ x=63.9835grams \\ \\ x\approx64\text{ }g \end{gathered}`

Hence, the mass in grams of O₂ that is used up in the reaction if a car produces 88 g of CO₂ is 64 g.

d. The grams of CO₂ and H₂O that can be produced if 125 g of C₂H₂O were burnt.

1 mole of C₂H₂O = 42.04 g

1 mole of CO₂ = 44.01 g

1 mole of H₂O = 18.01 g

Mass of CO₂ produced:

From the balanced, the mole ratio of C₂H₂O to CO₂ is 1:2

`\begin{gathered} 42.02g\text{ }C₂H₂O=2*44.01g\text{ }CO₂ \\ \\ 125g\text{ }C₂H₂O=x \\ \\ x=\frac{125g\text{ }C₂H₂O}{42.02g\text{ }C₂H₂O}*88.02g\text{ }CO₂ \\ \\ x\approx262g\text{ }CO₂ \end{gathered}`

The grams of CO₂ produced is 262 g.

Mass of H₂O produced:

From the balanced, the mole ratio of C₂H₂O to H₂O is 1:1

`\begin{gathered} 42.02g\text{ }C₂H₂O=18.01g\text{ }H₂O \\ \\ 125g\text{ }C₂H₂O=x \\ \\ x=\frac{125g\text{ }C₂H₂O}{42.02g\text{ }C₂H₂O}*18.01g\text{ }H₂O \\ \\ x\approx53.6grams \end{gathered}`

The mass of H₂O produced is 53.6 g