Another 10.00 g sample of the same hydrate is placed in a sealed 1.00 L flask wi

Question

Another 10.00 g sample of the same hydrate is placed in a sealed 1.00 L flask with air at 25.0 degree C and 741.0 torr. The flask is warmed to 100.0 degree C. The total pressure in the flask at this temperature is 1191.1 torr. Perform the following calculations. The reaction which occurs at this temperature is CaSO_4 middot 2 H_2O(s) rightarrow CaSO_4 middot 0.5 H_2O(s) + 1.5 H_2O(g) a. What is the partial pressure of the air at this temperature? b. What is the partial pressure of the water vapor at this temperature? c. How many moles of water are in the vapor phase at this temperature? d. What per cent of the original sample remains?

Explanation / Answer

From the ideal gas equation :

PV = nRT

at final conditions

T = 100C = 373 K

P = 1191.1 torr = 1.567 atm

R = 0.0.821

V= 1 L

n = 0.0511

Out of total moles ...according to equation

it produces 1.5 moles of water vapour and 1 mole of another one

moles of water vapour [ H2O gas] = 0.0511 * 1.5 / 3.5 = 0.0219

Moles of original = 0.0511 *1 / 3.5 = 0.0146

moles of another>

Partial pressure of water vapour = mole fraction * Total pressure = 510.47

Partial pressure of air = 1191.1 - 510.47 = 680.63 torr

Moles of water in vapour phase = 0.0219

Original sample % = 0.0146 / 0.0511 = 28.57

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