How many liters of hydrogen gas can be produced as 295k and 99.0 kPa if 22.3 grams of potassium metal react with excess water?

2K + H20 = 2KOH + H2

3 answers

You didn't balance the equation so I have done that.
2K + 2H20 = 2KOH + H2
mols K metal = grams/atomic mass = 22.3/39.1 = 0.570
The equation tells you with the coefficients aht 2 mols K will produce 1 mol H2 gas; therefore, 0.570 mols K will produce0.570/2 mols = ?
Then PV = nRT. You know P and n and R (8.314 if you use kPa), and T is 295 K. Solve for V in liters. Post your work if you get stuck
I think I have the correct answer, but would the answer be 6.50 L of hydrogen gas?
NOPE. I'll find the error if you show you set up and the math.