125g MgSO4 • 7 H2O how many grams of anhydrous compound can be obtained
Devron answered this for you below.
idek
3.51
125 x 0.02809 (7 mol MgSO4) = 3.51125 --> 3 significant figure = 3.51
To calculate the grams of anhydrous compound that can be obtained from 125g of MgSO4 • 7H2O, we need to account for the water molecules present in the hydrated compound.
First, let's determine the molar mass of MgSO4 • 7H2O:
- The molar mass of magnesium (Mg) is 24.31 g/mol.
- The molar mass of sulfur (S) is 32.07 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- The molar mass of hydrogen (H) is 1.01 g/mol.
For the anhydrous compound MgSO4, the molar mass is calculated as:
Mg: 1 atom x 24.31 g/mol = 24.31 g/mol
S: 1 atom x 32.07 g/mol = 32.07 g/mol
O: 4 atoms x 16.00 g/mol = 64.00 g/mol
Total: 120.38 g/mol
Now, let's calculate the grams of anhydrous compound using stoichiometry.
The molar mass of MgSO4 • 7H2O is 246.47 g/mol. This accounts for the additional water molecules in the compound.
To calculate the grams of anhydrous compound, use the following equation:
(125g MgSO4 • 7H2O) x (1 mol MgSO4 • 7H2O / 246.47 g MgSO4 • 7H2O) x (120.38 g MgSO4 / 1 mol MgSO4)
Performing the calculation:
(125g / 246.47) x (120.38 g) ≈ 61.05g of anhydrous compound.
Therefore, approximately 61.05 grams of anhydrous compound can be obtained from 125 grams of MgSO4 • 7H2O.