Respuesta :
Answer
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.
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:
[tex]C_2H_2O+2O_2\rightarrow2CO_2+H_2O[/tex]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;
[tex]\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}\times63.998g\text{ }O₂ \\ \\ x=63.9835grams \\ \\ x\approx64\text{ }g \end{gathered}[/tex]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
[tex]\begin{gathered} 42.02g\text{ }C₂H₂O=2\times44.01g\text{ }CO₂ \\ \\ 125g\text{ }C₂H₂O=x \\ \\ x=\frac{125g\text{ }C₂H₂O}{42.02g\text{ }C₂H₂O}\times88.02g\text{ }CO₂ \\ \\ x\approx262g\text{ }CO₂ \end{gathered}[/tex]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
[tex]\begin{gathered} 42.02g\text{ }C₂H₂O=18.01g\text{ }H₂O \\ \\ 125g\text{ }C₂H₂O=x \\ \\ x=\frac{125g\text{ }C₂H₂O}{42.02g\text{ }C₂H₂O}\times18.01g\text{ }H₂O \\ \\ x\approx53.6grams \end{gathered}[/tex]The mass of H₂O produced is 53.6 g
