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

Sorry i think the answer is b

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.