Calculate the maximum amount of product that can be formed and the amount of unreacted excess reagent when 3.1 mol of SO2 reacts with 2.7 mol of O2 according to the equation: 2SO2(g) + O2(g)->2SO3(g)
In this problem, we can solve this by comparing which of the reactants will produce the least amount of product.
`2 SO_2 (g) + O_2 (g) -> 2 SO_3 (g)`
So we will solve for the number of moles of `SO_3` produced using the two moles given.
`3.1 mol es SO_2 * (2 mol es SO_3)/(2 mol es SO_2) `
= 3.1 moles `SO_3`
`2.7 mol es O_2 * (2 mol es SO_3)/(1 mol es O_2) `
= 5.4 moles `SO_3`
We can see that the limiting reagent is `SO_2` and the excess reagent is the `O_2` and therefore we will use the amount of `SO_3` derived from the moles of `SO_2` .
From `SO_2` :
`3.1 mol es SO_3 *(80.0632 grams)/(1 mol e SO_3)` ` `
= 248.1959 = 248 grams `SO_3`
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