Acetylene (C2H2) and hydrogen fluoride (HF) react to give difluoroethane:

Question

C2H2+2HF->C2H4F2
When 2.3 mol of C2H2 and 12 mol HF are reacted in a 15.4 L flask, what will be the pressure in the flask at 13°C when the reaction is complete?

I know I'm supposed to find the limiting reagent, and use the formula PV=nRT; but I'm not clear on how to find the limiting reagent yet.

PLEASE HELP!!!

Answer 1

I find the limiting reagent by solving two simple stoichiometry problems. I use one reagent and find the product, then the other reagent and find the product. Convert 2.3 mol C2H4 to moles of the product.
2.3 mols C2H4 x (1 mole C2H4F2/1 mole C2H2) = 2.3 x (1/1) = 2.3 mols C2H4F2.

Then do the same for HF.
12 moles HF x (1 mole C2H4F2/2 moles HF) = 12 x (1/2) = 6 moles C2H4F2
Obviously both answers can't be right; the correct one in limiting reagent problems is ALWAYS the smaller one. In this case, the limiting reagent is C2H2 and there will be 2.3 moles C2H4F2 formed.