What is the limiting reagent in a reaction where 0.25L of 1.150 M HCl is used to neutralize 45g of Ba(OH)2?

Question

Here is the full equation:

2HCl(aq) + Ba(OH)2(aq) --> BaCl2(aq) +2H2O(l)

Thanks in advance for the help!

Answer 1

First find out how many moles of each reactant you have
For HCl: mole=M*V= 1.150*0.25 = 0.2875
For Ba(OH)2: mole= mass/MW (MW is the molecular weight) so
mole= 45/171 =0.2632

Now the stoichiometry of the reaction is
2 mole HCl react with 1 mole Ba(OH)2, thus
0.2875 mole HCl with x

so, 2x=1*0.2875 => x =0.2875/2 =0.1438 mole Ba(OH)2 are needed to react completely with the amount of HCl that you have.
But you have more Ba(OH)2 than that (0.2632 mole instead of 0.1438). Thus you have an excess of Ba(OH)2 and the limiting reagent is HCl.