Question

If 1.50 μg of CO and 6.80 μg of H2 were added to a reaction vessel, and the reaction went to completion, how many gas particles would there be in the reaction vessel assuming no gas particles dissolve into the methanol?

Answer 1

The chemical reaction would be as follows:

CO + 2H2 =CH3OH

We are given the amount of reactants to be used. We have to use these amounts to determine which is the limiting reactant and how much of the excess reactant is left.

1.50x10^-6 g CO ( 1 mol / 28.01 g ) = 5.36x10^-8 mol CO
6.80 x10^-6 g H2 ( 1 mol / 2.02 g ) = 3.37x10^-6 mol H2

Therefore, it is CO that is consumed completely in the reaction and the number of moles gas left would be 3.37x10^-6 - 5.36x10^-8 = 3.32x10^-6 moles.