Question

When CO2(g) (0.009242 mol/L) and 0.009242 mol/L of H2(g) in a 460.0 L reaction vessel at 619.0 °C are allowed to come to equilibrium the mixture contains 1.327 mol of CO(g). What is the equilibrium concentration (mol/L) of H2(g)?
CO2(g)+H2(g) = CO(g)+H2O(g)

Answer 1

This one is the same as the one just before and both use the same principle as all of the others.

Answer 2

is this simply x = -1.327

therefore .009242-(-1.327) = 1.3362

then 1.3362 / 460.00 = .0029?

Answer 3

Note here that CO and H2 are given in mols/L so you need to convert to mols first. Then it's done the same way as the CS2 and CH4 problem before.