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

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom