# 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?

## $S {O}_{2} + {O}_{2} \to S {O}_{3}$

Jun 15, 2016

$S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$

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.

$\text{Moles of sulfur dioxide} = \frac{90.0 \cdot g}{64.07 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.41 \cdot m o l$

$\text{Moles of dioxygen} = \frac{100.0 \cdot g}{32.02 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.12 \cdot m o l$

From the balanced equation, which unequivocally shows the stoichiometric relationship, the oxygen gas is in excess, and sulfur trioxide will react with $0.71 \cdot m o l$ dioxygen to form $1.41 \cdot m o l$ sulfur trioxide.

The excess dioxygen is simply: $\left(3.12 - 0.71\right) \cdot m o l$.