Answer:
See explanation
Explanation:
First, let's write the balanced equation again:
2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)
Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:
Ptotal = Pwat + PCO2 (1)
This is the dalton's law for partial pressures.
The pressure can be also be relationed with the moles
Ratio of mole = Ratio of pressure
so, taking this in consideration we can say the following:
Pwater/PCO2 = moles water / moles CO2
As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:
Pwater = Ptotal/2
Pwater = 7.76 / 2 = 3.88 atm
With this pressure, and using the ideal gas equation, we can know the moles of water:
PV = nRT
n = PV/RT using R = 0.082 L atm / K mol
n = 3.88 * 5 / 0.082 * (160+273)
n = 0.546 moles of water
b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.
2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g) MMCO2 = 84 g/mol
the moles of NaHCO3 initially:
n = 100 / 84
n = 1.19 moles
so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:
remaining moles = 1.19 - 0.546 = 0.644 moles
therefore the mass remaining:
mCO2 = 0.644 * 84
mCO2 = 54.096 g
c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:
Kp = Pwater * PCO2
Kp = 3.88 * 3.88
Kp = 15.0544
d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.
