# In the production of sulfuric acid, what is the limiting reagent when 1.30 moles of "S" react with 8.80 moles of "HNO"_3 ?

## $\text{S"(s) + 6"HNO"_3(aq) -> "H"_2"SO"_4(aq) + 6"NO"_2(g) + 2"H"_2"O} \left(l\right)$

Apr 2, 2018

Sulfur.

#### Explanation:

The balanced chemical equation that describes this reaction

${\text{S"_ ((s)) + 6"HNO"_ (3(aq)) -> "H"_ 2"SO"_ (4(aq)) + 6"NO"_ (2(g)) uarr + 2"H"_ 2"O}}_{\left(l\right)}$

tells you that every $1$ mole of sulfur that takes part in the reaction consumes $6$ moles of nitric acid, which means that sulfur and nitric acid take part in the reaction in a $1 : 6$ mole ratio.

In your case, you know that you start with $1.30$ moles of sulfur. According to the aforementioned mole ratio, your sample of sulfur will need

1.30 color(red)(cancel(color(black)("moles S"))) * "6 moles HNO"_3/(1color(red)(cancel(color(black)("mole S")))) = "7.80 moles HNO"_3

in order to be completely consumed by the reaction. To find the limiting reagent, all you need to do is to compare the number of moles of nitric acid that you have in your solution to the number of moles of nitric acid that you need in order for all the moles of sulfur to react.

Since you have a bigger number of moles of nitric acid than needed

overbrace("8.80 moles HNO"_3)^(color(blue)("what you have")) > overbrace("7.80 moles HNO"_3)^(color(blue)("what you need"))

you can say that nitric acid will be in excess. In other words, sulfur will be the limiting reagent because it will be completely consumed before all the moles of nitric acid present in the solution will get the chance to react.

So the reaction will consume $1.30$ moles of sulfur and $7.80$ moles of nitric acid and leave you with an excess of

${\text{8.80 moles " - " 7.80 moles" = "1.00 moles HNO}}_{3}$