H3PO4 + 3KOH = K3PO4 + 3 H2O What would be the percent yield if you reacted 49.0g of H3PO4 with 112g of potassium hydroxide, and collected 25.0g of water?

Answer 1

This is a limiting reagent problem (LR) and first you must determine what the theoretical yeild is for the reaction.
..........H3PO4 + 3KOH ==> K3PO4 + 3H2O
mols H3PO4 = grams/molar mass = approx 49/98 = 2
mols H2O produced if this is the LR.= 2*3 = 6
mols KOH = 112/56 = approx 2
mols H2O poduced if this is the LR = about 2
The smaller is the LR which is KOH so we should get
g H2O = mols H2O x molar mass H2O = 2 x 18 = 36 g.This is the theoretical yield = TY
You need to confirm all of these calculations INCLUDING the molar
masses.
The actualy yield (AY) from the problem si 25.0 g
%yield = [(AY/TY)]*100
Post your work if you get stuck.

Answer 2

just give the answer