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.

To determine the volume of oxygen needed at STP to completely react with 5.63 g of potassium, we need to use the balanced chemical equation:

4K + O2 -> 2K2O

Now let's follow the steps to solve the problem:

Step 1: Write the balanced equation.

The balanced equation is already written as 4K + O2 -> 2K2O.

Step 2: Convert the given mass of potassium into moles.

To do this, divide the given mass of potassium (5.63 g) by the molar mass of potassium (39.1 g/mol):

Moles of potassium (K) = 5.63 g / 39.1 g/mol = 0.144 mol

Step 3: Use the coefficients in the balanced equation to convert moles of potassium to moles of oxygen.

According to the balanced equation, for every 4 moles of potassium (K) reacted, 1 mole of oxygen (O2) is required.

Moles of oxygen (O2) = 0.144 mol K x (1 mol O2/4 mol K) = 0.036 mol

Step 4: Convert moles of oxygen to volume at STP.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Volume of oxygen (O2) at STP = 0.036 mol O2 x 22.4 L/mol = 0.8064 L

Therefore, the volume of oxygen needed at STP to completely react with 5.63 g of potassium is approximately 0.8064 liters.