One mole of carbon powder and 32 g of O2 gas are allowed to react in a closed 2.00-L flask. After the reaction is complete, and the temperature is allowed to equilibrate to 0 °C, what is the final pressure of CO2 in the flask?
Actually I don't get how to solve this question and what formula I have to use. I need a little Help, please.
Actually I don't get how to solve this question and what formula I have to use. I need a little Help, please.
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use molar mass, to find moles:
(32 g O2 used ) / (32.0 rams O2 / mole) = 1 .0 moles of O2 is being used
by the reaction:
1 mol C 1 mole O2 --> 1 Mole O2
we will make 1 mole of CO2 from the number of moles of O2 & C used
find pressure
PV = nRT
P (2.00L) = (1 mol) (0.08206 L-atm/mol-K) (273 Kelvin)
P = 11.2 atm of pressure
======================================…
or
since there is the Molar Volume ratio of 1 mole of gas has 22.4 litres at STP
& since STP is 0 Celsius & 1 atm
....
the relationship for 1 mole at 0 Celsius is also:
1 atm has 22.4 Litres
so
using Boyles law , that pressure is inversely proportional to volume:
P1V1 = P2V2
(1 atm) (22.4 L) = P2 (2 Litres)
P2 = (22.4 l) (1atm) / (2.00L)
P2 again = 11.2 atm
(32 g O2 used ) / (32.0 rams O2 / mole) = 1 .0 moles of O2 is being used
by the reaction:
1 mol C 1 mole O2 --> 1 Mole O2
we will make 1 mole of CO2 from the number of moles of O2 & C used
find pressure
PV = nRT
P (2.00L) = (1 mol) (0.08206 L-atm/mol-K) (273 Kelvin)
P = 11.2 atm of pressure
======================================…
or
since there is the Molar Volume ratio of 1 mole of gas has 22.4 litres at STP
& since STP is 0 Celsius & 1 atm
....
the relationship for 1 mole at 0 Celsius is also:
1 atm has 22.4 Litres
so
using Boyles law , that pressure is inversely proportional to volume:
P1V1 = P2V2
(1 atm) (22.4 L) = P2 (2 Litres)
P2 = (22.4 l) (1atm) / (2.00L)
P2 again = 11.2 atm