volume stoichiometry

Determine the volume of oxygen that is needed at STP to completely react with 5.63 g of potassium.

4K + O2 yields 2H2O

I don't buy the equation.
If K is reacting with oxygen, then the equation is
4K + O2 ==> 2K2O
Actually, I think it forms K2O2. Do you have any information in the problem that tells you the products. For example, I don't know how you ended up with H2O.

Step 1. Write the balanced equation. Done above.

Step 2. Convert what you have into mols.
mols = g/atomic mass = 5.63g/39.1 = ?? mols K on hand.

Step 3. Using the coefficients in the balanced equation from step 1, convert mols of what you have (in this case mols K) to mols of what you want (in this case mols oxygen).
mols O2 = ??mols K x (1 mol O2/4 mols K).
Note that the mols K (which you don't want) cancel to leave units of oxygen (which you want to keep).

step 4. Now convert mols oxygen to volume. You know 1 mol will occupy 22.4L at STP.
volume O2 at STP = mols O2 x 22.4 L/mol = xx liters O2.
This will give you the procedure, but unless you have been directed otherwise, I would change the equation at the start to form K2O2.
Post your work if you need further asistance.

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