Question

How many moles of barium hydroxide, `Ba(OH)_2` would be required to react with 117g hydrogen bromide, `HBr` in the reaction `2HBr + Ba(OH)_2 -> BaBr_2 + 2H_2O`?

Answer 1

Approx. 0.725 moles of `Ba(OH)_2`?

Explanation:

Moles of `HBr`? `=` `(117*g)/(80.9*g*mol^-1)` `=` `1.45*mol`.

The stoichiometry of the reaction explicitly tells us that each mole, each equivalent, of barium hydroxide requires 2 mole, 2 equivalents, of hydrogen bromide.

Thus `0.725` `mol` barium hydroxide mole are required.