How do you calculate the mass of excess reagent remaining at the end of the reaction in which 90.0g of #SO_2# are mixed with 100.0g of #O_2#?
#SO_2 + O_2-> SO_3#
You start with a balanced chemical equation:
It is very clear that dioxygen gas is in excess.
The balanced equation tell us that 64 g of sulfur dioxide react with 16 g of dioxygen to give 80 g of sulfur trioxide.
From the balanced equation, which unequivocally shows the stoichiometric relationship, the oxygen gas is in excess, and sulfur trioxide will react with
The excess dioxygen is simply: