A 0.940g sample of barium chloride dehydrate is heated and 0.800g of anhydrous residue remains after cooling.
A. How many moles of anhydrous barium chloride were present in the sample?
B. How many moles of water were present in the sample?
A.
(0.800g BaCl2) / (208.233g BaCl2/mol) =
0.00384 mol BaCl2
B.
(0.940 g - 0.800 g) /
(18.01532 g H2O/mol) = 0.00777 mol H2O
I don't answer the answer that's why I posted the question.
Well, let's break down the problem, molecule by molecule.
A. To find the moles of anhydrous barium chloride, we need to find the moles of the anhydrous residue that remains.
The molecular weight of barium chloride (BaCl2) is approximately 208.23 g/mol. So, for the 0.800g of anhydrous residue, we can use this formula:
moles = mass / molecular weight
Therefore, moles of barium chloride = 0.800g / 208.23 g/mol.
B. Now, let's find the moles of water (H2O) that were present in the sample.
The molecular weight of water is approximately 18.02 g/mol. Since we know the total mass of the sample (0.940g) and the mass of the anhydrous residue (0.800g), we can find the mass of the water that was lost.
Mass of water = total mass - mass of anhydrous residue
= 0.940g - 0.800g
Finally, we can find the moles of water using the formula:
moles = mass / molecular weight
So, moles of water = (0.940g - 0.800g) / 18.02 g/mol.
Now, don't go disappearing like that water!
To solve this problem, we need to use the molar masses of barium chloride and water. The molar mass of BaCl2 is 208.23 g/mol, and the molar mass of water is 18.015 g/mol.
A. To find the number of moles of anhydrous barium chloride, we need to calculate the mass of the anhydrous residue.
Mass of anhydrous barium chloride = Mass of sample - Mass of residue
= 0.940 g - 0.800 g
= 0.140 g
Now we can calculate the number of moles of anhydrous barium chloride:
Number of moles = Mass / molar mass
= 0.140 g / 208.23 g/mol
= 0.000672 mol
Therefore, there were 0.000672 moles of anhydrous barium chloride present in the sample.
B. To find the number of moles of water, we can use the mass of the sample and the mass of the anhydrous residue.
Mass of water = Mass of sample - Mass of anhydrous residue
= 0.940 g - 0.800 g
= 0.140 g
Now we can calculate the number of moles of water:
Number of moles = Mass / molar mass
= 0.140 g / 18.015 g/mol
= 0.007770 mol
Therefore, there were 0.007770 moles of water present in the sample.
To find the number of moles of anhydrous barium chloride present in the sample (Part A), we need to first calculate the number of moles of water that were eliminated during heating. Then, we can determine the number of moles of anhydrous residue that remains.
To find the number of moles of water in the sample (Part B), we need to calculate the difference between the total moles of compound and the moles of anhydrous residue.
Let's break down the steps:
A. Calculating the moles of anhydrous barium chloride:
1. Determine the molar mass of barium chloride dihydrate (BaCl2·2H2O):
- The atomic mass of barium (Ba) is 137.33 g/mol
- The atomic mass of chlorine (Cl) is 35.45 g/mol (x2 for the subscript 2)
- The atomic mass of hydrogen (H) is 1.008 g/mol (x2 for the subscript 2)
- The atomic mass of oxygen (O) is 16.00 g/mol (x2 for the subscript 2)
- The molar mass of BaCl2·2H2O = (137.33 g/mol) + (35.45 g/mol x 2) + (1.008 g/mol x 2) + (16.00 g/mol x 2) = 244.26 g/mol
2. Calculate the moles of barium chloride dihydrate (BaCl2·2H2O) present in the sample:
- Moles = Mass / Molar mass
- Moles = 0.940 g / 244.26 g/mol
B. Calculating the moles of water:
1. Determine the molar mass of water (H2O):
- The atomic mass of hydrogen (H) is 1.008 g/mol (x2 for the subscript 2)
- The atomic mass of oxygen (O) is 16.00 g/mol
- The molar mass of H2O = (1.008 g/mol x 2) + 16.00 g/mol = 18.02 g/mol
2. Calculate the moles of water present in the sample:
- Moles = Mass / Molar mass
- Moles = (0.940 g - 0.800 g) / 18.02 g/mol
Now you can substitute the values into the equations to find the answers.