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Solved: Ethanol (C2H5OH) burns in air: Balance the
Chapter 5, Problem 62P(choose chapter or problem)
Ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) burns in air:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
Balance the equation and determine the volume of air in liters at \(35.0^{\circ} \mathrm{C}\) and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent \(O_{2}\) by volume.
Questions & Answers
QUESTION:
Ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) burns in air:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
Balance the equation and determine the volume of air in liters at \(35.0^{\circ} \mathrm{C}\) and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent \(O_{2}\) by volume.
ANSWER:Step 1 of 3
The goal of the problem is to balance the equation and determine the volume of air in liters.
Given equation:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(1)\)
Given:
\(\mathrm{T}=35.0^{\circ} \mathrm{C}=35+273=308 \mathrm{~K}\)
\(\mathrm{P}=790 \mathrm{mmHg}\)
Mass of ethanol \(=227\)
Mass percent of \(\mathrm{O}_{2}\) in air \(=21.0 \%\)
The balanced equation is:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)