posted by andy .
When CO(g) (38.11 grams) and 96.48 grams of Cl2(g) in a 140.0 L reaction vessel at 782.0 K are allowed to come to equilibrium the mixture contains 0.4080 mol of Cl2CO(g). What is the equilibrium concentration (mol/L) of CO(g)?
CO(g)+Cl2(g) = Cl2CO(g) Can anyone help with the steps to how to complete this problem?
Set up an ICE chart.
moles CO = 38.11/molar mass CO = ??
moles COCl2 at equilibrium = 0.4080; therefore, COCl2 must have increased by 0.4080 mols while CO an CL2 decreased by 0.4080. How much does that leave for CO.
Initial moles - moles at equilibrium = moles remaining. That divided by 140 L = concn.