I have a chemistry question. I know a lot of people will say do your hw, but I have tried. A person can only do so much by themself. I need help. That said.... how would you do this problem?
A 5.0g of dioxane(C4H8O2) is placed in a 2.0L, air-tight steel container filled with pure oxygen at
1atm and 28°C. The sample is then combusted and allowed to cool to 28°C. Approximate the
final pressure of the container.
A 5.0g of dioxane(C4H8O2) is placed in a 2.0L, air-tight steel container filled with pure oxygen at
1atm and 28°C. The sample is then combusted and allowed to cool to 28°C. Approximate the
final pressure of the container.
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PV = nRT
n = PV / RT = (1 atm) x (2.0 L) / ((0.08205746 L atm/ K mol) x (28 + 273K)) = 0.080974 mol O2 initially
(5.0 g C4H8O2) / (88.1056 g C4H8O2/mol) = 0.056750 mol C4H8O2
C4H8O2 + 5 O2 → 4 CO2 + 4 H2O
0.080974 mole of O2 would react completely with 0.080974 x (1/5) = 0.0161948 mole of C4H8O2, but there is more C4H8O2 present than that, so C4H8O2 is in excess and O2 is the limiting reactant.
(As a practical matter, when O2 is the limiting reactant, there will be some products of incomplete combustion such as CO and C as well as CO2, but the present problem doesn't provide enough information to deal with such possibilities, so complete combustion is still assumed.)
Supposing the left over dioxane to be liquid and of negligible volume, likewise the water produced.
(2.0 L O2) x (4/5) = 1.6 L CO2 produced
(1 atm) x (1.6 / 2.0) = 0.8 atm of CO2, which is the final pressure, more or less, since a lot of dubious assumptions have been made
n = PV / RT = (1 atm) x (2.0 L) / ((0.08205746 L atm/ K mol) x (28 + 273K)) = 0.080974 mol O2 initially
(5.0 g C4H8O2) / (88.1056 g C4H8O2/mol) = 0.056750 mol C4H8O2
C4H8O2 + 5 O2 → 4 CO2 + 4 H2O
0.080974 mole of O2 would react completely with 0.080974 x (1/5) = 0.0161948 mole of C4H8O2, but there is more C4H8O2 present than that, so C4H8O2 is in excess and O2 is the limiting reactant.
(As a practical matter, when O2 is the limiting reactant, there will be some products of incomplete combustion such as CO and C as well as CO2, but the present problem doesn't provide enough information to deal with such possibilities, so complete combustion is still assumed.)
Supposing the left over dioxane to be liquid and of negligible volume, likewise the water produced.
(2.0 L O2) x (4/5) = 1.6 L CO2 produced
(1 atm) x (1.6 / 2.0) = 0.8 atm of CO2, which is the final pressure, more or less, since a lot of dubious assumptions have been made