Final answer:
To determine the mass of 3.6 moles of Mg(NO3)2, calculate its molar mass (148.313 g/mol) and multiply by the number of moles, resulting in 533.9268 grams.
Step-by-step explanation:
To calculate the mass of 3.6 moles of magnesium nitrate, Mg(NO3)2, first we need to determine the molar mass of the compound. The molar mass is calculated by summing the atomic masses of all the atoms in a formula unit of the compound. Magnesium (Mg) has an atomic mass of 24.305 g/mol, nitrogen (N) has an atomic mass of 14.007 g/mol, and oxygen (O) has an atomic mass of 15.999 g/mol. The compound contains one Mg atom, two N atoms, and six O atoms, so its molar mass is:
- 1 Mg = 24.305 g/mol
- 2 N = 2 × 14.007 g/mol = 28.014 g/mol
- 6 O = 6 × 15.999 g/mol = 95.994 g/mol
Add these together to get the molar mass:
24.305 g/mol + 28.014 g/mol + 95.994 g/mol = 148.313 g/mol
Now, to find the mass in grams of 3.6 moles of Mg(NO3)2:
3.6 moles × 148.313 g/mol = 533.9268 grams
Therefore, you would expect to have 533.9268 grams of Mg(NO3)2 if you had 3.6 moles of the substance.