convert the reaction to moles.
each mole occupies 22.4L at STP, so adjust that volume using PV=kT
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?
3 answers
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.
you balanced this wrong lol
3 answers