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Sagot :
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
Read more at; https://brainly.com/question/10594788
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