Question

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?

Answer 1

To determine the number of gas particles in the vessel we add all of the components of the gas. For this, we need to convert the mass to moles by the molar mass. Then, from moles to molecules by the avogadro's number.

1.50x10^-6 ( 1 / 28.01) (6.022x10^23) = 3.22x10^16 molecules CO


6.80x10^-6 ( 1 / 2.02) (6.022x10^23) = 2.03x10 18 molecules H2

Totol gas particles = 2.05x10^18 molecules

Answer 2

Answer: The number of gas particles remained in the vessel is $13.81\times 10^{17}$

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For CO:

Given mass of CO = $1.50\mu g=1.50\times 10^{-6}g$       (Conversion factor:  $1\mu g=10^{-6}g$  )

Molar mass of CO = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of CO}=\frac{1.50\times 10^{-6}g}{28g/mol}=0.053\times 10^{-6}mol$

• For hydrogen gas:

Given mass of hydrogen gas = $6.8\mu g=6.8\times 10^{-6}g$

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{6.8\times 10^{-6}g}{2g/mol}=2.4\times 10^{-6}mol$

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

$CO+2H_2\rightarrow CH_3OH$

By Stoichiometry of the reaction:

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

So, $0.053\times 10^{-6}$ moles of carbon monoxide will react with = $\frac{2}{1}\times 0.053\times 10^{-6}=0.106\times 10^{-6}mol$ 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.

• Amount of excess reagent (hydrogen gas) left = $(2.4-0.106)\times 10^{-6}=2.294\times 10^{-6}$ moles

According to mole concept:

1 mole of an element or compound contains $6.022\time 10^{23}$  number of particles.

So, $2.294\times 10^{-6}moll$ of hydrogen gas will contain = $2.294\times 10^{-6}\times 6.022\times 10^{23}=13.81\times 10^{17}$ number of particles.

Hence, the number of gas particles remained in the vessel is $13.81\times 10^{17}$