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

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

b) The FORMULA for the limiting reagent is

c) Mass of excess reagent remains is 4.8 grams

Explanation:

To calculate the moles :

According to stoichiometry :

1 mole of require = 2 moles of

Thus 0.196 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent as (0.467-0.392) = 0.075 moles or are left.  

As 1 mole of give = 2 moles of

Thus 0.196 moles of give =  of

Mass of

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

6.4 cm (max) - 25 cm (min)