Question

If 11.3 g of sulfur reacts with 14.0 g of oxygen, how many grams of sulfur trioxide could be formed? (S + O2 → SO₃)

a) 25.3 g

b) 22.6 g

c) 19.2 g

d) 15.8 g

Answer 1

To find the grams of sulfur trioxide formed, convert the given masses of sulfur and oxygen to moles. Compare the moles of sulfur and oxygen to determine the limiting reactant. None of the option is applicable.

Step-by-step explanation:

To find the grams of sulfur trioxide formed, we need to determine the limiting reactant in the reaction.

First, we convert the given masses of sulfur and oxygen to moles using their molar masses (Sulfur: 1 mole = 32.07 g, Oxygen: 1 mole = 16.00 g). Sulfur: 11.3 g / 32.07 g/mol = 0.352 mol. Oxygen: 14.0 g / 16.00 g/mol = 0.875 mol.

Next, we compare the moles of sulfur and oxygen to the stoichiometric ratio in the balanced equation: 1 mole of sulfur reacts with 1 mole of oxygen to form 1 mole of sulfur trioxide (SO₃). Since we have an excess of oxygen (0.875 mol > 0.352 mol), sulfur is the limiting reactant.

Therefore, the moles of sulfur trioxide formed is also 0.352 mol. Finally, we convert the moles of sulfur trioxide to grams using the molar mass of sulfur trioxide (1 mole = 80.06 g/mol): 0.352 mol x 80.06 g/mol = 28.18 g.