Question

one litre of oxygen at stp is made to react with three times the volume of CO at stp . Find the composition of the resultant mixture .

Answer 1

since ... CO + 0.5O2 = CO2

we can see here that 0.5 moles of O2 reacts with 1 mole of CO ...

now 1ltr of O2 contains ..... 1/22.4   moles

it needs 1/11.2 moles of CO.....
so only 2 litres of CO would get used and 1ltr remains......

and reaction gives 2 litres of CO2
so now composition of mixture is ....

1ltr CO and 2ltr od CO2

Answer 2

Answer: The volume of the product obtained will be 2 L.

Explanation:

According to the mole concept:

1 mole of a gas occupies 22.4L of volume.

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

$CO+\frac{1}{2}O_2\rightarrow CO_2$

1 mole of carbon monoxide will occupy 22.4 L of volume.

And, $\frac{1}{2}$ moles of oxygen gas will occupy $\frac{1}{2}\times 22.4=11.2L$ of volume.

By Stoichiometry of the reaction:

11.2 L of oxygen gas reacts with 22.4 L of carbon monoxide gas.

So, 1L of oxygen gas reacts with $\frac{22.4}{11.2}\times 1=2L$ of carbon monoxide gas.

As, the given amount of carbon monoxide is more than the required amount. Thus, it is considered as an excess reagent.

Oxygen gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

11.2 L of oxygen gas produces 22.4 L of carbon dioxide gas.

So, 1L of oxygen gas produces $\frac{22.4}{11.2}\times 1=2L$ of carbon dioxide gas.

Hence, the volume of the product obtained will be 2 L.