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?

The chemical reaction would be written as:

CO + 2H2 = CH3OH

We need to determine first the limiting reactant from the given amounts of the reactants. Then, use this amount to relate the substances from the reaction. We calculate as follows:

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

Therefore, the limiting reactant would be CO.

H2 left = 3.37x10^-6 mol H2 - (5.36x10^-8)(2) = 3.2628x10^-6 H2 left

The correct answer is c