Answer:
We can for 5.93 grams potassium fluoride
Explanation:
Step 1: Data given
Mass of potassium = 4.00 grams
Mass of fluorine = 3.00 grams
Molar mass potassium = 39.10 g/mol
Molar mass fluorine gas =38.00 g/mol
Step 2: The balanced equation
2K (s) + F2 (g) → 2KF (s)
Step 3: Calculate moles potassium
Moles potassium = 4.00 grams / 39.10 g/mol
Moles potassium = 0.102 moles
Step 4: Calculate moles F2
Moles F2 = 3.00 grams / 38.00 g/mol
Moles F2 = 0.0789 moles
Step 5: Calculate limiting reactant
Potassium is the limiting reactant. There will react 0.102 moles
Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles
There will remain 0.0789 - 0.051 = 0.0279 moles
Step 6: Calculate moles potassium fluoride
For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride
For 0.102 moles K we need 0.102 moles KF
Step 7: Calculate mass KF
Mass KF = moles KF * molar mass KF
Mass KF = 0.102 moles * 58.10 g/mol
Mass KF = 5.93 grams
