How many sulfur atoms are present in 0.25 mol of Al2S3?
3 * .25 * 6 * 10^23
To find the number of sulfur atoms present in 0.25 mol of Al2S3, we need to consider the chemical formula and use Avogadro's number.
1. Start by writing the chemical formula for aluminum sulfide: Al2S3. This means that each molecule of Al2S3 contains 2 aluminum atoms (Al) and 3 sulfur atoms (S).
2. Determine the molar mass of Al2S3 by calculating the sum of the atomic masses of its constituent elements. The atomic mass of aluminum (Al) is approximately 26.98 g/mol, and the atomic mass of sulfur (S) is approximately 32.06 g/mol. Therefore,
molar mass of Al2S3 = (2 * atomic mass of Al) + (3 * atomic mass of S)
= (2 * 26.98 g/mol) + (3 * 32.06 g/mol)
= 53.96 g/mol + 96.18 g/mol
= 150.14 g/mol
3. Determine the number of moles of Al2S3 given in the question. We are given that the amount is 0.25 mol.
4. Use Avogadro's number to find the number of formula units in 0.25 mol of Al2S3. Avogadro's number is approximately 6.022 x 10^23 formula units per mole.
number of formula units = number of moles * Avogadro's number
= 0.25 mol * (6.022 x 10^23 formula units/mol)
= 1.5055 x 10^23 formula units
5. Finally, multiply the number of formula units by the number of sulfur atoms per formula unit to find the total number of sulfur atoms.
number of sulfur atoms = number of formula units * number of sulfur atoms per formula unit
= (1.5055 x 10^23 formula units) * (3 sulfur atoms per formula unit)
= 4.5165 x 10^23 sulfur atoms
Therefore, there are approximately 4.5165 x 10^23 sulfur atoms in 0.25 mol of Al2S3.