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#

1 Answer
anor277
Jun 15, 2016

You start with a balanced chemical equation:
#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#

It is very clear that dioxygen gas is in excess.

Explanation:

The balanced equation tell us that 64 g of sulfur dioxide react with 16 g of dioxygen to give 80 g of sulfur trioxide.

#"Moles of sulfur dioxide"=(90.0*g)/(64.07*g*mol^-1)# #=# #1.41*mol#

#"Moles of dioxygen"=(100.0*g)/(32.02*g*mol^-1)# #=# #3.12*mol#

From the balanced equation, which unequivocally shows the stoichiometric relationship, the oxygen gas is in excess, and sulfur trioxide will react with #0.71*mol# dioxygen to form #1.41*mol# sulfur trioxide.

The excess dioxygen is simply: #(3.12-0.71)*mol#.

Related questions
See all questions in Limiting Reagent
Impact of this question
23289 views around the world
You can reuse this answer
Creative Commons License