Answer:
14.3 g SO₃
Explanation:
2S + 3O₂ → 2SO₃
First, find the limiting reactant. To do that, calculate the mass of oxygen needed to react with all the sulfur.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂
0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂
There are 10.0 g of O₂, so there's enough oxygen. The limiting reactant is therefore sulfur.
Use the mass of sulfur to calculate the mass of sulfur trioxide.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃
0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃
