When SbCl3(g) (0.001539 mol/L) and 0.6772 mol of Cl2(g) in a 440.0 L reaction vessel at...

When SbCl3(g) (0.001539 mol/L) and 0.6772 mol of Cl2(g) in a 440.0 L reaction vessel at 665.0 K are allowed to come to equilibrium the mixture contains 0.0009506 mol/L of SbCl5(g). What is the concentration (mol/L) of SbCl3(g)? SbCl3(g)+Cl2(g) = SbCl5(g)

Expert Solution

SbCl₃(g) + Cl₂(g) = SbCl₅(g)

t=0 0.001539 mol/L 0.6772/440 mol/l 0

t = teq 0.001539 -x 0.001539 -x x = 0.0009506 mol/L given

[ SbCl₃(g) ] = 0.001539 - 0.0009506 = 5.884 x 10^-4 mol/l

queen_honey_blossom answered 9 months ago

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