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

Is there a diagram/picture you can provide? it seems this question goes along with a

Ch4 is nonpolar and insoluble in water. since all of the hydrogen atoms are evenly spread around the carbon atom even through the c h bond is polar all the poles cancel each other out.