How many grams of potassium fluoride can form if 4.00 grams of potassium are reacted with 3.00 grams of fluorine gas according to the reaction: 2K (s) + F2 (g) → 2KF (s)

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

it is stable as it is in group 0 which have full outer shells of electrons