Determine the value of x in al2(so4)3 . xh2o when 4 grams of aluminum sulfate hydrate was heated and 3.5 grams remain. (show working please)
To determine the value of x in Al2(SO4)3 . xH2O, we need to determine the amount of water that was lost during the heating process.
First, let's calculate the number of moles for both aluminum sulfate and water at the start:
1. Calculate the molar mass of Al2(SO4)3:
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- So, the molar mass of Al2(SO4)3 is (2*26.98) + (3*32.07) + (12*16.00) = 342.12 g/mol.
2. Calculate the number of moles of Al2(SO4)3 at the start:
- Number of moles = mass / molar mass
- Number of moles = 4 g / 342.12 g/mol = 0.0117 mol.
3. Determine the number of moles of water initially present:
- Since the formula is Al2(SO4)3 . xH2O, we need to calculate the ratio of moles of water to moles of aluminum sulfate.
- From the balanced equation of Al2(SO4)3, we see that there are 6 moles of water for every 1 mole of Al2(SO4)3.
- So, the number of moles of water initially present = 0.0117 mol * 6 = 0.0702 mol.
Now, let's calculate the number of moles of water remaining:
1. Calculate the molar mass of H2O:
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- So, the molar mass of H2O is 1.01 + (2 * 16.00) = 18.02 g/mol.
2. Calculate the mass of water remaining:
- Mass of water remaining = mass at the start - mass remaining (as given in the question)
- Mass of water remaining = 4 g - 3.5 g = 0.5 g.
3. Calculate the number of moles of water remaining:
- Number of moles = mass / molar mass
- Number of moles = 0.5 g / 18.02 g/mol = 0.0277 mol.
Finally, determine the value of x:
1. Calculate the moles of water lost:
- Moles of water lost = moles of water initially present - moles of water remaining
- Moles of water lost = 0.0702 mol - 0.0277 mol = 0.0425 mol.
2. Calculate the value of x:
- Since there are 6 moles of water for every 1 mole of Al2(SO4)3, we can set up the proportion:
(0.0702 mol of water) / (1 mol of Al2(SO4)3) = (0.0425 mol of water lost) / (x mol of Al2(SO4)3)
- Solving for x, we get x = (0.0425 mol * 1 mol of Al2(SO4)3) / (0.0702 mol) ≈ 0.606
Therefore, the value of x in Al2(SO4)3 . xH2O is approximately 0.606.