2H2O(l)→2H2(g)+O2(g) What mass of H2O is required to form 1.6 L of O2 at a temperature of 310 K and a pressure of 0.902 atm ?
- electron1Lv 72 years ago
According to the coefficients in the balanced equation, two moles of water will produce two moles of hydrogen and one mole of oxygen. At standard temperature and pressure, the volume of one mole of a gas is 22.4 liters. Let’s use the following equation to determine the volume of one mole of oxygen at standard temperature and pressure.
P1 * V1 ÷ T1 = P2 * V2 ÷ T2
0.906 * 1.6 ÷ 320 = 1 * V2 ÷ 273
V2 * 320 = 395.7408
V2 = 395.7408 ÷ 320 = 1.23669 liters
To determine the number of moles of oxygen, divide this volume by 22.4.
n = 1.23669 ÷ 22.4
This is approximately 0.055 mole. According to the coefficients in the balanced equation, the number of moles of water is twice this number.
n = 1.23669 ÷ 11.2
The mass of one mole of water is 18 grams.
Mass = 18 * (1.23669 ÷ 11.2)
This is approximately 2 grams. I like using this method to solve this type of problem, because I did not have to use rounded numbers to get the final answer. I hope this is helpful for you.
- Roger the MoleLv 72 years ago
n = PV / RT = (0.902 atm) x (1.6 L) / ((0.08205746 L atm/K mol) x (310 K)) = 0.056734 mol O2
(0.056734 mol O2) x (2 mol H2O / 1 mol O2) x (18.01532 g H2O/mol) = 2.04 g H2O
- billrussell42Lv 72 years ago
2H₂O (l) → 2H₂ (g) + O₂ (g)
or 2 moles of H₂O yields 2 moles of H₂ + 1 mole of O₂
or 36 g of H₂O yields 4 g of H₂ + 32 of O₂
1.6 L of O₂ at 210 K and 0.902 atm
n = PV/RT = (0.902)(1.6) / (0.08206)(210) = 0.0837 mole
2 moles of H₂O to get 1 mole of O₂
therefore 2• 0.0837 moles of H₂O to get 0.0523 mole of O₂
or 0.1674 mol
H₂O is 18 g/mol
18 g/mol x 0.1674 mol = 3.012 g
Ideal gas law
PV = nRT
n = number of moles
R = gas constant = 0.08206 (atm∙L)/(mol∙K)
T = temperature in kelvins
P = absolute pressure in atm
V = volume in liters