Question

when one mole of SO2 and 1 mole of O2 are made to react completion
(1)All the oxygen will be consumed
(2)1.0 mole of SO3 will be produced
(3)0.5 mole of SO2 is remained
(4)All of these

Answer 1

To check my approach and convention, check out my other answer in the given link: https://brainly.in/question/4163747
The balanced reaction is:
$2SO_2+O_2\longrightarrow 2SO_3\\\dfrac{n_{SO_2}}{v_{SO2}}=\dfrac{1}{2}\\\dfrac{n_{O_2}}{v_{O_2}}=\dfrac{1}{1}=1\\\Rightarrow \dfrac{n_{SO_2}}{v_{SO2}}<\dfrac{n_{O_2}}{v_{O_2}}\\\Rightarrow SO_2 \text{ is the limiting reagent.}\\\Rightarrow \dfrac{n_{SO_2}}{v_{SO_2}}=\dfrac{n_{SO_3}}{v_{SO_3}}\\\Rightarrow \dfrac{1}{2}=\dfrac{n_{SO_3}}{2}\\\Rightarrow n_{SO_3}=1$
So option (2) is correct.

Answer 2

Answer:

When one mole of SO2 and 1 mole of O2 are made to react completion 1.0 mole of SO3 will be produced.

Explanation:

$2SO_{2}$ + $O_{2}$ → $2SO_{3}$

Finding the limiting reagent,

$SO_{2} = \frac{given moles}{stiochiometric cofficient} = \frac{1}{2} = 0.5$

$O_{2}= \frac{given moles}{stiochiometric coffiecent}= \frac{1}{1}= 1$

Since the fraction for  $SO_{2}$ is less, therefore $SO_{2}$ will be the limiting reagent that is it gets completely consumed.

We know that the amount of product formed will be decided by the amount of limiting reagent.

2 moles of $SO_{2}$ ⇒ 2 moles of $SO_{3}$.

1 mole of $SO_{2}$ ⇒ 1 mole of $SO_{3}$.

$O_{2}$ is present in excess.

1 mole of $SO_{2}$ will only reacts with 0.5 moles $O_{2}$.

Thus, 0.5 mol of $O_{2}$ will left unreacted.

#SPJ2