415 of anhydrous barium chloride when 459g of the hydrated salt were heated.calculate n in the formula Bacl2 nh2o
I need the answer now please
the molar mass of BaCl2 is 208 g
the moles (BaCl2) in the heated sample is ... 415 / 208
the mass lost by heating is H2O ... 459 g - 415 g
... find the moles by dividing by the molar mass of H2O
n is ... (moles H2O) / (moles BaCl2)
To calculate the value of 'n' in the formula BaCl2·nH2O, where 'n' represents the number of water molecules, we need to use the given information and the molar mass of the compounds involved.
First, we need to find the mass of water lost. The mass lost during heating is the difference between the mass of the hydrated salt and the mass of the anhydrous salt.
Mass of hydrated salt = 459 g
Mass of anhydrous salt = 415 g
Mass of water lost = Mass of hydrated salt - Mass of anhydrous salt
= 459 g - 415 g
= 44 g
Next, we need to calculate the number of moles of water lost. To do this, we divide the mass of water lost by the molar mass of water.
Molar mass of water (H2O) = 18 g/mol
Number of moles of water lost = Mass of water lost / Molar mass of water
= 44 g / 18 g/mol
≈ 2.44 mol
Finally, we can determine the value of 'n' by comparing the ratio of moles between the anhydrous salt and water. From the chemical formula BaCl2·nH2O, we know that for each mole of anhydrous salt, 'n' moles of water are present.
Moles of anhydrous salt = Moles of water lost
1 mol BaCl2 = 2.44 mol water
From this ratio, we can conclude that 'n = 2.44'. Therefore, the formula for the hydrated barium chloride is BaCl2·2H2O.
To determine the value of "n" in the formula BaCl2 · nH2O, we need to use the given information and apply the concept of mole ratio in a chemical reaction.
The molar mass of anhydrous barium chloride (BaCl2) is the sum of the molar masses of barium (Ba) and chlorine (Cl), which are 137.33 g/mol and 35.45 g/mol, respectively. Therefore, the molar mass of BaCl2 is 137.33 + (2 × 35.45) = 208.23 g/mol.
Similarly, the molar mass of water (H2O) is 2 × 1.01 + 16.00 = 18.02 g/mol.
We know that the mass of the hydrated salt (BaCl2 · nH2O) is 459 g and that the mass of the anhydrous salt (BaCl2) is 415 g.
To find the number of moles of anhydrous barium chloride, we divide the mass by the molar mass:
moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2
moles of BaCl2 = 415 g / 208.23 g/mol
moles of BaCl2 = 1.994 mol
Now, to find the number of moles of water, we subtract the number of moles of anhydrous BaCl2 from the number of moles of the hydrated salt:
moles of H2O = moles of hydrated salt - moles of BaCl2
moles of H2O = 459 g / 18.02 g/mol - 1.994 mol
moles of H2O = 25.47 mol - 1.994 mol
moles of H2O = 23.476 mol
Since each formula unit of BaCl2 · nH2O contains "n" moles of water, we can set up a ratio based on the number of moles we calculated:
n = moles of H2O / moles of BaCl2
n = 23.476 mol / 1.994 mol
n ≈ 11.78
Therefore, the value of "n" in the formula BaCl2 · nH2O is approximately 11.78.