250.0 mL of 0.800 M HBr and 150.0 mL of 1.00 M LiOH are mixed together.
a) write balanced equation
b) determine the moles of both reagents
c) which reagent is the limiting reagent?
d) label acid and base as strong or weak
e) calculate how many grams of the salt being produced

Question

250.0 mL of 0.800 M HBr and 150.0 mL of 1.00 M LiOH are mixed together.
a) write balanced equation
b) determine the moles of both reagents 
c) which reagent is the limiting reagent?
d) label acid and base as strong or weak
e) calculate how many grams of the salt being produced

anonymous · Answer

a) LiOH + HBr --> LiBr + H20

B/C) Using molarity and number of liters convert to mols, there are .28 mols of HBr and .15 mol LiOH.  From the balanced equation you see that it is a 1 to 1 mol ratio therefore LiOH is the limiting reagent

d) HBr is a strong acid, LiOH is a strong base

e) 13.02 g of salt produced

cometailcane · Answer

thank you!

anonymous · Answer

No problem.  Let me know if you need me to explain anything.