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 number of gas particles remained in the vessel is

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of CO =       (Conversion factor:    )

Molar mass of CO = 28 g/mol

Putting values in equation 1, we get:

Given mass of hydrogen gas =

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of carbon monoxide and hydrogen gas follows:

By Stoichiometry of the reaction:

1 mole of carbon monoxide reacts with 2 moles of hydrogen gas.

So, moles of carbon monoxide will react with = of hydrogen gas.

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon monoxide is considered as a limiting reagent because it limits the formation of product. and it is completely consumed in the reaction.

According to mole concept:

1 mole of an element or compound contains  number of particles.

So, of hydrogen gas will contain = number of particles.

Hence, the number of gas particles remained in the vessel is

16

explanation:

the marks go by 4s

gas

explanation:

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