mass SbCl5 = 89.7 g
molar mass SbCl5 = 299.0 g/mol
moles SbCl5 = 0.3 mol
molarity SbCl5 = 0.02 mol/L = 0.02 M
The concentration of SbCl5 is 0.02 mol/L.
Ideal Gas Equation:
V = 15.0 L
T = 182°C + 273.15 = 455.15 K
P = 0.747 atm
At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas, as shown by the following equation:
SbCl5 ⇌ SbCl3 + Cl2
An 89.7-gram sample of SbCl5 (molecular weight = 299.0) is placed in an evacuated 15.0-liter container at 182°C.
(1) What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
(2) What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?
b) If the SbCl5 is 29.2% decomposes when equilibrium is established at 182°C, calculate the value for either equilibrium constant, Kp or Kc, for this decomposition reaction. Indicate whether you are calculating Kp or Kc.
c) In order to produce some SbCl5, a 1.00-mole sample of SbCl3 is first placed in an empty 2.00-liter container maintained at a temperature different from 182°C. At this temperature, Kc equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?
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