Question

Calculate the mass of So3 (g) produced, if 500 g so2, (g) reacts with 200 g 02 (g)
according to the equation : 2SO2(g) + O2(g) → 2S03().
Identify the limiting reagent.​

Answer 1

Answer: 625 grams and $SO_2$ is the limiting reagent .

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $SO_2$

$\text{Number of moles}=\frac{500g}{64g/mol}=7.81moles$

b) moles of $O_2$

$\text{Number of moles}=\frac{200g}{32g/mol}=6.25moles$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require 1 mole of $O_2$

Thus 7.81 moles of $SO_2$ require=$\frac{1}{2}\times 7.81=3.91moles$  of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product.

As 2 moles of $SO_2$ give =  2 moles of $SO_3$

Thus 7.81 moles of $SO_2$ give =$\frac{2}{2}\times 7.81=7.81moles$  of $SO_3$

Mass of $SO_3=moles\times {\text {Molar mass}}=7.81moles\times 80g/mol=625g$

Thus mass of $SO_3$ produced is 625 grams and $SO_2$ is the limiting reagent .