2KOH + H2SO4 ==> K2SO4 + 2H2O
Just another LR problem. You can ALWAYS spot LR problems because amounts for BOTH reactants are given in the problem. If just one reactants is given (and usually it say an excess of the other one is used) that is a simple stoichiometry problem like the Zn/HCl problem below.
Write the balanced neutralization reaction between H2SO4 and KOH in aq. solution.
.350 L of .410 M H2SO4 is mixed with .300 L of .230 M KOH. What concentration of sulfuric acid remains after neutralization?
______ M H2SO4
6 answers
The limiting reactant is the KOH with .069 moles right? So then I go from mol KOH to grams of K2SO4?
Yes, KOH is the limiting reagent. There's a new twist here. They don't ask for mols or grams of one of the products; instead they ask for how much H2SO4 is left. That's done the same way.
Convert mols KOH used (0.069) to mols H2SO4 used, subtract from H2SO4 there initially and that gives you mols H2SO4 left. Then mols H2SO4/L solution = M. Don't forget, total volume will be L KOH + L H2SO4.
Convert mols KOH used (0.069) to mols H2SO4 used, subtract from H2SO4 there initially and that gives you mols H2SO4 left. Then mols H2SO4/L solution = M. Don't forget, total volume will be L KOH + L H2SO4.
.156 M?
Oh, Got it! .168?
I don't get either of those.
1.435 = mols H2SO4 initially
-0.069/2 = mols H2SO4 used
1.400 = mols H2SO4 left
Then M = mols/L = 1.400/0.65L = about 2.15M
1.435 = mols H2SO4 initially
-0.069/2 = mols H2SO4 used
1.400 = mols H2SO4 left
Then M = mols/L = 1.400/0.65L = about 2.15M
1 answer