How many total moles of ions are released when each of the following samples dissolves completely in water?
30.2g of Ba(OH)2 8H2O
Ba(OH)2.8H2O ==>Ba^+2 + 2OH^- + 8H2O
How many mols of the compound did you start with initially. That will be 30.2 g Ba(OH)2.H2O/molar mass of that material.
Then you get 1 mol Ba^+, + 2 mols OH^- + 8 mols H2O. Add them to get total mols. The question may or may not be a trick question. Note that it asks for the total moles of IONS. If that is the intent of the problem, then the 8 mols H2O aren't included in the total since they are not ions.
Well, I have to admit, moles don't typically enjoy getting wet. But in this case, we'll make an exception. Let's dive into the aquatic world of chemistry!
Ba(OH)2•8H2O is a hydrate, containing 8 water molecules for every Ba(OH)2 formula unit. So first, we need to calculate the number of moles in 30.2g of Ba(OH)2 • 8H2O.
The molar mass of Ba(OH)2 • 8H2O can be calculated as follows:
(1 Ba * atomic mass Ba) + (2 O * atomic mass O) + (16 H * atomic mass H) + (8 H2O * molar mass H2O)
Now the atomic mass of Ba, O, and H can be found on the periodic table, while the molar mass of H2O is approximately 18 g/mol.
So the total molar mass of Ba(OH)2 • 8H2O can be calculated. And by dividing the given mass of 30.2g by the molar mass, we can find the number of moles.
Finally, since Ba(OH)2 dissociates in water to form 3 ions per formula unit (1 Ba2+ ion and 2 OH- ions), we can multiply the number of moles by 3 to find the total moles of ions released.
But hey, I'm just a funny bot, not a calculator. So grab your periodic table and give those calculations a go!
To determine the total moles of ions released when 30.2g of Ba(OH)2 8H2O dissolves completely in water, we need to consider the dissociation of the compound.
First, let's calculate the molar mass of Ba(OH)2 8H2O:
Molar mass of Ba(OH)2 = 137.33 g/mol
Molar mass of H2O = 18.02 g/mol
The mass of 8H2O can be calculated as:
Mass of H2O = 8 * 18.02 g/mol = 144.16 g/mol
Therefore, the molar mass of Ba(OH)2 8H2O is:
Molar mass of Ba(OH)2 8H2O = 137.33 g/mol + 144.16 g/mol = 281.49 g/mol
Now, let's calculate the moles of Ba(OH)2 8H2O:
Moles of Ba(OH)2 8H2O = Mass / Molar mass
Moles of Ba(OH)2 8H2O = 30.2 g / 281.49 g/mol ≈ 0.1073 mol
Ba(OH)2 dissociates into Ba^2+ and 2 OH^- ions when it dissolves in water.
Therefore, the total moles of ions released can be calculated as:
Total moles of ions = Moles of Ba^2+ + (Moles of OH^- × 2)
Moles of Ba^2+ = Moles of Ba(OH)2 8H2O = 0.1073 mol
Moles of OH^- = 2 × Moles of Ba(OH)2 8H2O = 2 × 0.1073 mol = 0.2146 mol
Total moles of ions = 0.1073 mol + (0.2146 mol × 2) = 0.5365 mol
Therefore, when 30.2g of Ba(OH)2 8H2O dissolves completely in water, a total of approximately 0.5365 moles of ions are released.
To find the number of moles of ions released when the given sample dissolves completely in water, we need to first determine the number of moles of the compound.
Step 1: Calculate the molar mass of Ba(OH)2 · 8H2O
Ba(OH)2·8H2O consists of a barium ion (Ba2+) and two hydroxide ions (OH-) along with 8 water molecules (H2O).
The molar masses are:
Ba: 137.33 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
To calculate the molar mass of Ba(OH)2 · 8H2O:
Ba = 1 × 137.33 g/mol = 137.33 g/mol
O = 10 × 16.00 g/mol = 160.00 g/mol (8 O from OH- and 2 O from H2O)
H = 20 × 1.01 g/mol = 20.20 g/mol (16 H from H2O and 4 H from OH-)
Total molar mass = 137.33 g/mol + 160.00 g/mol + 20.20 g/mol = 317.53 g/mol
Step 2: Convert grams to moles
Using the molar mass calculated in step 1, we can convert the given mass of the compound to moles.
moles = mass / molar mass
moles of Ba(OH)2 · 8H2O = 30.2 g / 317.53 g/mol ≈ 0.095 mol
Step 3: Identify the moles of ions released
In Ba(OH)2, there is one Ba2+ ion and two OH- ions.
Therefore, when 0.095 mol of Ba(OH)2 · 8H2O dissolves completely in water, the total moles of ions released are:
Ba2+ ions = 0.095 mol × 1 = 0.095 mol
OH- ions = 0.095 mol × 2 = 0.190 mol
So, when 30.2 g of Ba(OH)2 · 8H2O dissolves completely in water, there are 0.095 moles of Ba2+ ions and 0.190 moles of OH- ions released.