Asked by Jade
1.86 g of ethanol reacts with 10.0 g of oxygen. What is the total volume of gas present (in L) after the reaction is complete, assuming the reaction takes place at 1.00 atm and 25oC?
Answers
Answered by
oobleck
convert the reaction to moles.
each mole occupies 22.4L at STP, so adjust that volume using PV=kT
each mole occupies 22.4L at STP, so adjust that volume using PV=kT
Answered by
DrBob222
The question asks for the TOTAL volume of gas which will be the CO2 produced PLUS the oxygen remaining that did not react.
CH3COOH + 2O2 ==> 2CO2 + 2H2O
mols CH3COOH = grams/molar mass = 1.86/60 = 0.031
That will use 0.031 x 2 = 0.062 mols O2.
mols O2 initially = 10.0 g/32 = 0.312
mols O2 unreacted = 0.312 - 0.062 = 0.25
mols CO2 formed = 0.031 x 2 = 0.062
The water @ 25 C will not be a gas.
Total mols gas = 0.25 mols O2 + 0.062 mols CO2 = 0.312 @ 273 K. You want the volume @ 298 K so use PV = nRT
Post your work if you get stuck.
CH3COOH + 2O2 ==> 2CO2 + 2H2O
mols CH3COOH = grams/molar mass = 1.86/60 = 0.031
That will use 0.031 x 2 = 0.062 mols O2.
mols O2 initially = 10.0 g/32 = 0.312
mols O2 unreacted = 0.312 - 0.062 = 0.25
mols CO2 formed = 0.031 x 2 = 0.062
The water @ 25 C will not be a gas.
Total mols gas = 0.25 mols O2 + 0.062 mols CO2 = 0.312 @ 273 K. You want the volume @ 298 K so use PV = nRT
Post your work if you get stuck.
Answered by
muffin
you balanced this wrong lol
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