To calculate the mass of \(0.150 \, \text{mol}\) of barium hydroxide (\(Ba(OH)_2\)), you need to know its molar mass. Let's break it down:
Barium (\(Ba\)) has a molar mass of approximately \(137.33 \, \text{g/mol}\).
Oxygen (\(O\)) has a molar mass of approximately \(16.00 \, \text{g/mol}\).
Hydrogen (\(H\)) has a molar mass of approximately \(1.01 \, \text{g/mol}\).
So, for \(Ba(OH)_2\):
\[Molar \, mass = (1 \times M_{Ba}) + (2 \times M_{O}) + (2 \times M_{H})\]
\[= (1 \times 137.33) + (2 \times 16.00) + (2 \times 1.01)\]
\[= 137.33 + 32.00 + 2.02\]
\[= 171.35 \, \text{g/mol}\]
Now, to find the mass of \(0.150 \, \text{mol}\) of \(Ba(OH)_2\), we use the formula:
\[ Mass = \text{Molar mass} \times \text{Number of moles} \]
\[ = 171.35 \, \text{g/mol} \times 0.150 \, \text{mol} \]
\[ = 25.70 \, \text{g} \]
So, \(0.150 \, \text{mol}\) of \(Ba(OH)_2\) is \(25.70 \, \text{g}\).