Stoichiometry Problem

Question

LiOH + HBr → LiBr + H2O

-If you start with 10.0 grams of lithium hydroxide, how many grams of lithium bromide will theoretically be produced?
-How many water molecules will you produce if you start with 15.5 grams of lithium hydroxide
-If you start with 11.0 grams of lithium hydroxide, how many moles of lithium bromide should be produce if there is 100% yield?

Answer 1

First, check that the equation is balanced.  It is (but always check.  Never assume).

 

• Calculate # moles of LiOH:  Using the stoichiometry, and dimensional analysis, 
• 10 g LiOH x 1 mole LiOH/23.93 g LiOH = 0.418 moles LiOH
• Again, using the stoichiometry:
• 0.418 moles LiOH x (1 mole LiBr/mole LiOH) = 0.418 moles LiBr produced theoretically
• Using the stoichiometry again & Avogadro's number -
• 15.5 g LiOH x (1 mole LiOH / 23.94 g LiOH) = 0.647 moles LiOH
• 0.647 moles LiOH x (1 mole H₂O / mole LiOH) x (6.023 x 10²³ molecules H₂O/mole H₂O) =           3.897 x 10²³ molecules of H₂O   produced from 15.5 g LiOH
• Again - calculate moles LiOH in 11.0 g, and convert to moles of LiBr at 100% yield:
• 11.0 g LiOH x (1 mole LiOH/23.94 g LiOH) x (1 mole LiBr/mole LiOH) = 0.459 moles LiBr at 100% yield