Posted by **ANNA-MARIE** on Tuesday, April 19, 2011 at 6:08pm.

When SbCl3(g) (4.867 mol) and 0.01217 mol/L of Cl2(g) in a 400.0 L reaction vessel at 958.0 K are allowed to come to equilibrium the mixture contains 0.008586 mol/L of SbCl5(g). What concentration (mol/L) of SbCl3(g) reacted?

SbCl3(g)+Cl2(g) = SbCl5(g)

Unit Conversions:

K = C + 273

Molar Mass (g/mol)

SbCl3(g) 228.16

Cl2(g) 70.906

SbCl5(g) 299.06

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