Question

For the following reaction, 29.9 grams of sulfur dioxide are allowed to react with 6.26 grams of oxygen gas . sulfur dioxide(g) + oxygen(g) sulfur trioxide(g) What is the maximum mass of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer: a)  The maximum mass of sulfur trioxide that can be formed is 31.4 grams

b) The FORMULA for the limiting reagent is $O_2$

c) Mass of excess reagent remains is 4.8 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles$

$\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles$

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

According to stoichiometry :

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

Thus 0.196 moles of $O_2$ will require=$\frac{2}{1}\times 0.196=0.392moles$  of $SO_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $SO_2$ is the excess reagent as (0.467-0.392) = 0.075 moles or $0.075mol\times 64g/mol=4.8g$ are left.  

As 1 mole of $O_2$ give = 2 moles of $SO_3$

Thus 0.196 moles of $O_2$ give =$\frac{2}{1}\times 0.196=0.392moles$  of $SO_3$

Mass of $SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g$

Thus 31.4 g of $SO_3$ will be produced from the given masses of both reactants.