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 182oC. 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

x = = 0.50 M

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 182oC. 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

Moles Cl2 that must be added = 0.40 mol

Explanation:

K = ( = 0.117

Equilibrium concentrations:

= (1.00 - 0.70) mol / 2.00 L = 0.15 M <<---why is it 1.00-0.70???

= 0.700 mole / 2.00L = 0.350 M

= x

Kc = ÷ (0.15) = 0.117

Moles Cl2 at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol

Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol

Moles Cl2 that must be added = 0.40 mol