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How many liters of oxygen gas can be produced if 28.7 grams of water decomposes at 294 Kelvin and 0.986 atmospheres? Show all of the work used to solve this problem. . . 2 H2O (l) --->2 H2 (g) + O2 (g)

Answer :

2 H₂O (l) → 2 H₂ (g) + O₂ (g) 

Molar mass water = 1.01 x 2 + 16.00 = 18.02 g/mol 

Number of mol water decomposed = 28.7 g H₂O x [1 mol / 18.02g] = 1.59 mol H₂O 

From the balanced equation 2 mol H₂O decomposes to 2 mol H₂ and 1 mol O₂ 

so the mole ratio water : oxygen = 2 : 1 

and number of mol O₂ produced = ½ x 1.59 = 0.796 mol O₂ 


The ideal gas law is PV = nRT 

so V = nRT/P 

P = 0.986 atm 

V = ? 

n = 0.796 

R = 0.0821 L atm K⁻¹ mol⁻¹ 

T = 294K 

V = 0.796 x 0.0821 x 294 / 0.986 

V = 19.5 L 

So 19.5 L O₂ gas are produced

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