In the first reaction, sulfur is the limiting reactant and will produce 15.8 gms of sulfur trioxide gas. Thus, options a, and c are correct.
Limiting reactants can be given as the reactant species whose concentration determines the rate of the reaction, and the product formation.
The chemical reaction is shown as:
2S + 3O₂ → 2SO₃
Moles of sulfur: 6.3 ÷ 32.065 = 0.197 moles
Moles of oxygen: 10 ÷ 16 = 0.625 moles
From the reaction, it can be said that moles of sulfur are less than the oxygen and hence will limit the product formation.
The mass of the sulfur trioxide gas will be estimated by the moles of sulfur.
If 2 moles of sulfur = 2 moles of sulfur trioxide gas, then 0.197 moles of sulfur trioxide gas will be produced.
Mass of sulfur trioxide = moles × molar mass
= 0.197 moles × 80.06 g/mol
= 15.77 gms.
Therefore, 15.77 gm is the mass of sulfur trioxide gas.
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