Question

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`

Answer 1

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`.