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Fittoniya [83]
3 years ago
11

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)
Chemistry
1 answer:
Shtirlitz [24]3 years ago
5 0
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 

<span>From the balanced equation 2 mol H₂O decomposes to 2 mol H₂ and 1 mol O₂ </span>

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 

<span>So 19.5 L O₂ gas are produced

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

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