Calculate NaHCO3 Molar Mass & Equilibrium Constants

[gtfish]

Here's the question (ANY tips or help would be great):

Given equation- 2NaHCO3(s)<-->Na2CO3(s)+H2O(g)+CO2(g)
a. A sample of 100 grams of solid NaHCO3 was placed in an 5 L container and heated to 160 degrees C. Some of the original solid remained and the total pressure in the container was 7.76 atmospheres when equilibrium was reached. Calculate the number of moles of H2O present at equilibrium. (NaHCO3 molar mass= 84 g, PV=nRT, R=.08206 L atm/mol K)
b. How many grams of the original solid remain in the container under the conditions described in "a"?
c.Write the equilibrium expression for the equilibrium constant Kp and calculate its value for the reaction under the conditions in "a".
d. If 115 g. of solid NaHCO3 had been placed in the 5 L container and heated to 160 degrees C, what would have the total pressure been at equilibrium?

 

[himanshu121]

The Total Pressure will be due to water vapours+CO₂
The individual pressure will be 7.76/2 . so u have partial pressure of H₂O U have its Volume And Temperature.
So u can calculate the no of mol of Water vapours by the eqn u have mentioned

For part (b)

u have initial no of mol of NaHCO₃ as 100/84

final no of mol= initial - 2*(no of mol of H2O) [? Why answer urself]

I believe it is better that u do the rest of the Parts/ Or show what u have tried so far on the problem

 

[ravenprp]

a. To calculate the number of moles of H2O present at equilibrium, we can use the ideal gas law equation PV=nRT. We know that the pressure (P) is 7.76 atm, the volume (V) is 5 L, the temperature (T) is 160 degrees C or 433 K (since we are using the ideal gas law, we need to convert the temperature to Kelvin), and the gas constant (R) is 0.08206 L atm/mol K. We can rearrange the equation to solve for n, the number of moles of H2O: n=(PV)/(RT)=(7.76 atm x 5 L)/(0.08206 L atm/mol K x 433 K)=0.089 moles of H2O.

b. To calculate the number of grams of the original solid remaining, we can use the molar mass of NaHCO3 to convert moles to grams. We know that the original sample was 100 grams, and we calculated in part a that 0.089 moles of H2O were present at equilibrium. Using the molar ratio from the balanced equation, we can calculate the moles of NaHCO3 that reacted: 0.089 moles of H2O x (2 moles of NaHCO3/1 mole of H2O)=0.178 moles of NaHCO3 reacted. Therefore, the remaining moles of NaHCO3 are 100 g - (0.178 moles x 84 g/mol)=85.49 g of NaHCO3.

c. The equilibrium expression for Kp is Kp=[(P(H2O) x P(CO2))/P(NaHCO3)]^2. Using the values from part a, we can plug in the pressure values for H2O, CO2, and NaHCO3: Kp=[(7.76 atm x 7.76 atm)/(7.76 atm)]^2=60.25.

d. If 115 g of solid NaHCO3 were placed in the 5 L container and heated to 160 degrees C, the total pressure at equilibrium can be calculated using the same method as in part a. We know that the number of moles of NaHCO3 is 115 g/84 g/mol=1.369 moles. Using the molar ratio