# 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?

Apr 10, 2016

Approx. 0.725 moles of $B a {\left(O H\right)}_{2}$?
Moles of $H B r$? $=$ $\frac{117 \cdot g}{80.9 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.45 \cdot m o l$.
Thus $0.725$ $m o l$ barium hydroxide mole are required.