What mass of oxygen would be required to convert 7.00g of sulfur into sulfur trioxide

Answer 1

Step 1 - Write a balanced chemical reaction equation

Since oxygen gas appears naturally as O₂, start with that. O₂ + S → SO₃. To balance this, since there is and odd number of oxygen atoms on the right side, add a 2 prefix in front of the SO₃ resulting in: O₂ + S → 2SO₃. Now, add prefixes to the left side of the equation to balance it resulting in:

3O₂ + 2S → 2SO₃

Step 2 - Convert the 7g of sulfur into moles

We do this because we are going to use the chemical reaction equation to tell us how much oxygen is required but the chemical reaction equation is in units of moles not grams.

To convert to moles, we use the molar mass. The molar mass of sulfur is 32.066 g/mol. To convert, we divide 7g by 32.066 g/mol to get 0.2183 moles of S

Step 3 - Use the molar ratios to calculate moles of oxygen

The chemical reaction equation tells us it takes 3 moles of O₂ for every 2 moles of S so we must multiply the 0.2183 moles of S by 3/2 (moles O₂/moles S) so that results in 0.32745 moles O₂

Step 4 - Convert moles of O₂ to grams of O₂

To convert to grams, we use the molar mass. The molar mass of O₂ is 2x15.999 or 31.998 g/mol. To convert, we multiply 0.32745 moles O₂ by 31.998 g/mol to get 10.478 grams of O₂. There are 3 significant figures in the original given information so we round the answer to 3 sig figs or 10.5 g O₂