Question
Suppose you are going to add 450 mL of a 0.10 mol/L hydrochloric acid solution to 250 mL of a 0.2 mol/L sodium hydroxide solution.
a)Which one is the limiting reactant?
b)What is the theoretical yield of sodium chloride?
a)Which one is the limiting reactant?
b)What is the theoretical yield of sodium chloride?
Answers
DrBob222
These react 1:1; i.e.,
NaOH + HCl ==> H2O + NaCl
So which has the fewer moles? moles = M x L = ?
Use that substance to determine how much NaCl is formed.
NaOH + HCl ==> H2O + NaCl
So which has the fewer moles? moles = M x L = ?
Use that substance to determine how much NaCl is formed.
Maame
i don't understand
Maame
i would be great to show me the answer , but do it step by step so i understand better
DrBob222
Did you try it? I gave you the first step and you didn't show you did anything with it except to look at it and say "I don't understand."
moles NaOH = 0.2M x 0.250L = 0.050.
moles HCl = 0.1M x 0.450L = 0.045
.........NaOH + HCl ==> NaCl + H2O
initial..0.05...0.045....0.......0
So the 0.045 moles HCl will be used up completely leaving 0.050-0.045 NaOH = 0.005 moles NaOH remaining unreacted. moles NaCl formed = 0.045.
g NaOH = moles NaOH x molar mass NaOH.
And the limiting reagent is .....(the one that's completely used up).
moles NaOH = 0.2M x 0.250L = 0.050.
moles HCl = 0.1M x 0.450L = 0.045
.........NaOH + HCl ==> NaCl + H2O
initial..0.05...0.045....0.......0
So the 0.045 moles HCl will be used up completely leaving 0.050-0.045 NaOH = 0.005 moles NaOH remaining unreacted. moles NaCl formed = 0.045.
g NaOH = moles NaOH x molar mass NaOH.
And the limiting reagent is .....(the one that's completely used up).