Question

Sulfur reacts with oxygen to produce sulfur trioxide gas. If 3.15 g of sulfur reacts with 5 g of oxygen, what is the limiting reactant? How many grams of sulfur trioxide will be produced?

Im really having a horrible time in chemistry right now and just do not understand stoichiometry at all! Please help!

Answer 1

We address the stoichiometric equation.....

Explanation:

`underbrace(S(s) + 3/2O_2(g))_"80 g of reactant" rarr underbrace(SO_3(g))_"80 g of product"`

And as written, the reaction SPECIFICALLY invokes stoichiometry. Mass is conserved in every chemical reaction, and thus, if there are `80*g` of reactant, THERE MUST be `80*g` of product. And this notion of stoichiometry, i.e. `"garbage out EQUALS garbage in"` conforms ABSOLUTELY to EACH and EVERY chemical reaction EVER studied.

And again, I have written here before that we use this notion of stoichiometry every time we make a cash or electronic transaction. For every debit item, there must be a corresponding CREDIT item. And if stoichiometry is not observed here, we very soon notice the shortfall, the short change, or the over change. We are trained to recognize this from a very early age with money. We can apply this training and write it large when we deal with molar quantities of atoms and molecules, so long as we know the molar quantities of atoms and molecules (and we does!). Here endeth the diatribe.

And so here, we determine the molar quantities of each reactant:

`"Moles of sulfur"=(3.15*g)/(32.06*g*mol^-1)=0.0983*mol`.

`"Moles of dioxygen"=(5.00*g)/(32.00*g*mol^-1)=0.156*mol`.

And, here, clearly, there is a stoichiometric excess of dioxygen gas. Do you agree? Please don't trust my arifmetick.

And so here, sulfur is the limiting reagent. At most, we can make `0.0983*mol` of `SO_3`....i.e. a mass of `0.0983*molxx80.06*g*mol^-1=9.44*g`.

Anyway, if this isn't clear raise an objection or query, and someone will help you.