Asked by SteveJanker
How many liters of hydrogen gas can be produced at 300 K and 104 kPa pressure if 25.5 g of sodium metal reacts?
2Na + 2H20 = 2NaOH + H2
2Na + 2H20 = 2NaOH + H2
Answers
Answered by
oobleck
25.5g of Na = 1.108 moles
The equation says that you will get 0.554 moles of H2
Now you want V such that
104V/300 = 101.325 (0.554*22.4) / 273
The equation says that you will get 0.554 moles of H2
Now you want V such that
104V/300 = 101.325 (0.554*22.4) / 273
Answered by
Anonymous
Na = 23 grams / mol
how many mols of Na
25.5 g Na (1 mol/23g) = 1.11 mols of Na
so 1.11/2 = 0.554 mols of H2
P V = n R T
104*10^3 Pascals * n liters = 0.554 * R * 300
alternately
or one mol is 22.4 liter at 273 K and 1 atm or about 10*5 Pascals
so 0.554 mols is 12.4 liters at 273 K and 1 atm
P1 V1/T1 = P2 V2/T2
10^5 *12.4 / 273 = 1.04*10^5 V2 / 300
V2 is about 13 liters :)
how many mols of Na
25.5 g Na (1 mol/23g) = 1.11 mols of Na
so 1.11/2 = 0.554 mols of H2
P V = n R T
104*10^3 Pascals * n liters = 0.554 * R * 300
alternately
or one mol is 22.4 liter at 273 K and 1 atm or about 10*5 Pascals
so 0.554 mols is 12.4 liters at 273 K and 1 atm
P1 V1/T1 = P2 V2/T2
10^5 *12.4 / 273 = 1.04*10^5 V2 / 300
V2 is about 13 liters :)
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