Elemental sulfur occurs as octatomic molecules, S8. What mass of fluorine gas is needed for complete reaction with 24.1 g sulfur to form sulfur hexafluoride? I don't know where to even start with this question
To determine the mass of fluorine gas needed for the reaction, you need to follow several steps:
Step 1: Write a balanced chemical equation for the reaction.
The chemical equation for the reaction between sulfur and fluorine to form sulfur hexafluoride is:
S8 + 24 F2 → 8 SF6
Step 2: Calculate the molar mass of sulfur (S).
Consult the periodic table to find the atomic mass of sulfur, which is approximately 32.07 g/mol.
Step 3: Convert the given mass of sulfur (24.1 g) to moles.
Use the molar mass of sulfur to convert grams to moles:
mol of S = mass of S / molar mass of S
mol of S = 24.1 g / 32.07 g/mol ≈ 0.751 mol
Step 4: Determine the mole ratio between sulfur and fluorine.
By examining the balanced chemical equation, you can see that for every 1 mole of sulfur (S8), 24 moles of fluorine (F2) are required.
Step 5: Calculate the moles of fluorine needed.
Multiply the moles of sulfur by the mole ratio from the previous step:
mol of F2 = mole ratio × mol of S
mol of F2 = 24 × 0.751 mol ≈ 18.03 mol
Step 6: Convert the moles of fluorine to grams.
Use the molar mass of fluorine (F2), which is approximately 38.00 g/mol, to convert moles to grams:
mass of F2 = moles of F2 × molar mass of F2
mass of F2 = 18.03 mol × 38.00 g/mol ≈ 685.14 g
Therefore, approximately 685.14 grams of fluorine gas is needed for the complete reaction with 24.1 g of sulfur to form sulfur hexafluoride.