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.

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