holaa Posted May 10, 2023 Report Share Posted May 10, 2023 5 dm^{3} of carbon monoxide, CO(g), and 2 dm^{3} of oxygen, O_{2}(g), at the same temperature and pressure are mixed together. Assuming complete reaction according to the equation given, what is the maximum volume of carbon dioxide, CO_{2}(g), in dm^{3}, that can be formed? 2CO(g) + O_{2}(g) → 2CO_{2}(g) The answer is 4, but I don't know how to get there, can someone please explain this. Am I supposed to change the dm3 to moles? Reply Link to post Share on other sites More sharing options...

IB_taking_over Posted May 20, 2023 Report Share Posted May 20, 2023 You don't need to convert to moles. It is basically like fairly simple algebra equation. (There may be a faster way to do this; It's been a few years since I've done anything chemistry related.) We know that 2CO(g) + O_{2}(g) → 2CO_{2}(g) which means it takes 2 CO and 1 O2 to make 2 CO2. We have 5 dm^{3} of CO(g), and 2 dm^{3} of O_{2}(g) And we want to know much CO_{2}(g), in dm^{3 }these will produce. Since the units are all the same we don't need to convert anything. So, Step 1 - find the limiting reagent. The O2 can make 2 CO2 and the CO2 can make 2.5 CO2. So O2 is the limiting reagent. Step 2 - multiply the reaction equation by the amount the limiting reagent can produce. 2(2CO(g) + 1O_{2}(g) → 2CO_{2}(g)) We get: 4CO(g) + 2O_{2}(g) → 4CO_{2}(g) So, 4 dm3 CO2 Hopefully this make sense. If not hopefully someone else can explain it better 🙂 Reply Link to post Share on other sites More sharing options...

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