# How many moles are in 363.8 g of Ba(OH)_2?

Nov 30, 2015

$2.1 \text{mol } B a {\left(O H\right)}_{2}$

#### Explanation:

Determine the molar mass of the barium hydroxide.

Ba=(137.327"g")/"mol"

O=(16"g")/"mol"

H=(1.008"g")/"mol"

molar mass of Ba(OH)_2=((137.327+2(16)+2(1.008))"g")/"mol"

=(171.343"g")/"mol"

We know we have $363.8 \text{g}$.

$\left(363.8 \text{g")/(x" mol")=(171.343"g")/(1"mol}\right)$

Solve to find that there are $2.1 \text{mol } B a {\left(O H\right)}_{2}$.