Question

60. The following quantities are placed in a container: 1.5 X 1024 atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen. (a) What is the total mass in grams for the collection of all three elements? (b) What is the total number of moles of atoms for the three elements? (c) If the mixture of the three elements formed a compound with molecules that contain two hydrogen atoms, one sulfur atom, and four oxygen atoms, which substance is consumed first? (d) How many atoms of each remaining element would remain unreacted in the change described in (c)?

Answer 1

(a). The mass of hydrogen = (1.5*10²⁴/6.023*10²³) * 2 g/mol = 4.98 g

The mass of sulfur = 1 mol * 32 g/mol = 32 g

The mass of O2 = 88 g

The total mass = 4.98+32+88 = 124.98

(b) The moles of atoms of H = (1.5*10²⁴/6.023*10²³) = 2.49 mol

The moles of atoms of S = 1 mol

The moles of atoms of O₂ = (88/32)*2 = 5.5 mol

Therefore, the no. of moles of atoms = 4.98 + 1 + 5.5 = 11.48 mol

(c) The limiting reactant is sulfur, in the formation of the molecule of H2SO4, hence sulfur is consumed first.

(d) The remaining atoms of H = (2.49-2)*6.023*10²³ = 2.95*10²³ H-atoms

The remaining atoms of O = (5.5-4)*6.023*10²³ = 9.03*10²³ O-atoms