Asked by Anonymous
How many liters of oxygen are necessary to burn 30.4 liters of ethyl alcohol (C2H5OH)?
(Assume complete combustion)
(Assume complete combustion)
Answers
Answered by
DrBob222
C2H5OH + 3O2 ==> 2CO2 + 3H2O
You will need the density of ethyl alcohol. Use density to convert 30.4L to grams. mass = volume x density.
The convert g EtOH to mols. mols = g/molar mass.
Using the coefficients in the balanced equation, convert moles EtOH to moles O2.
moles O2 = moles EtOH x (1 mol O2/1 mol EtOH) = moles EtOH x (1/1) = ?
Then moles O2 x 22.4L/mol = L O2 required.
Post your work if you get stuck.
You will need the density of ethyl alcohol. Use density to convert 30.4L to grams. mass = volume x density.
The convert g EtOH to mols. mols = g/molar mass.
Using the coefficients in the balanced equation, convert moles EtOH to moles O2.
moles O2 = moles EtOH x (1 mol O2/1 mol EtOH) = moles EtOH x (1/1) = ?
Then moles O2 x 22.4L/mol = L O2 required.
Post your work if you get stuck.
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