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)
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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.
Using `SO_2`
`3.1 mol es SO_2 * (2 mol es SO_3)/(2 mol es SO_2) `
= 3.1 moles `SO_3`
Using `O_2`
`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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limiting is SO2 because you need 2 moles of SO2 for every 1 mole of O2. So you have to set up a mole to mole ratio, and you'll find that you need 1.55 moles of O2 to react with 3.1 moles of SO2 (SO2 is limiting because you still have some O2 remaining). Then you set up another mole to mole ratio with the product and you should get the maximum yield to be 3.1 moles of SO3.
MATH WORK: 3.1(1/2)=1.55 moles O2
1.55(2/1)=3.1 moles SO3
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