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

Answer: The number of gas particles remained in the vessel is
`13.81* 10^(17)`

Step-by-step 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* 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}=(1.50* 10^(-6)g)/(28g/mol)=0.053* 10^(-6)mol`

• For hydrogen gas:

Given mass of hydrogen gas =
`6.8\mu g=6.8* 10^(-6)g`

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

`\text{Moles of hydrogen gas}=(6.8* 10^(-6)g)/(2g/mol)=2.4* 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* 10^(-6)` moles of carbon monoxide will react with =
`(2)/(1)* 0.053* 10^(-6)=0.106* 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)* 10^(-6)=2.294* 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* 10^(-6)moll` of hydrogen gas will contain =
`2.294* 10^(-6)* 6.022* 10^(23)=13.81* 10^(17)` number of particles.

Hence, the number of gas particles remained in the vessel is
`13.81* 10^(17)`

Answer 2

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