Asked by Jimmy
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
2K + H20 = 2KOH + H2
Answers
Answered by
DrBob222
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
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
Answered by
Jimmy
I think I have the correct answer, but would the answer be 6.50 L of hydrogen gas?
Answered by
DrBob222
NOPE. I'll find the error if you show you set up and the math.
There are no AI answers yet. The ability to request AI answers is coming soon!