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what volume of gas is generated when 58.0 l of oxygen gas reacts at stp according to the following balanced equation? assume that the reaction goes to completion and that no starting materials remain. ch3ch2oh (l) + 3o2 (g) → 2co2 (g) + 3h2o (l)

Respuesta :

The volume of gas generated when 58.0 L of oxygen gas reacts at STP according to the given chemical equation reaction is;

Volume of CO₂ = 38.67 L

  • We want to find the volume of gas generated STP(standard temperature and pressure).

At STP, Temperature is 0°C and Pressure is 1 atm.

1 mole of gas will occupy a volume of 22.4 L.

  • We are told that the reaction goes to completion and as such, the balanced equation of the reaction is;

CH₃CH₂OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 3H₂O (l)

From this reaction, we can see that 3 moles of oxygen gas (O₂) produces 2 moles of Carbon (IV) Oxide (CO₂). The mole ratio of O₂ to CO₂ is;  

3:2

  • Since we are told that 58 L of oxygen reacts at STP, the using that ratio, we can find the corresponding volume of CO₂ produced as;

Volume of CO₂ = 58  × ²/₃

Volume of CO₂ = 38.67 L

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