!AXEL!
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Report Thread starter 11 months ago
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hi, i am doing a chem paper and this question always gets me. pls can you explain to me how it works i have a vague idea but it doesn't always work

8.0dm3 of NO is mixed with 6.0dm3 of O2 at room temperature and pressure (RTP).
The reaction below takes place until one of the reactants is used up.
2NO(g) + O2(g) → 2NO2(g)
What is the volume of the mixture at RTP after the reaction has taken place?
A 8.0dm3
B 10.0dm3
C 12.0dm3
D 14.0dm3
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pastelsun
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(Original post by !AXEL!)
hi, i am doing a chem paper and this question always gets me. pls can you explain to me how it works i have a vague idea but it doesn't always work

8.0dm3 of NO is mixed with 6.0dm3 of O2 at room temperature and pressure (RTP).
The reaction below takes place until one of the reactants is used up.
2NO(g) + O2(g) → 2NO2(g)
What is the volume of the mixture at RTP after the reaction has taken place?
A 8.0dm3
B 10.0dm3
C 12.0dm3
D 14.0dm3
As the reaction takes places at RTP, you can use the equation: moles x 24 = volume in dm3
Use that to find the number of moles of NO and O2 that are used in the reaction
Then identify which one of the reactants is the limiting reagent
The moles of NO2 depends on the number of moles of the limiting reagent, so use that to identify the moles of NO2
Then use the equation to work out the volume
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Pigster
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(Original post by !AXEL!)
hi, i am doing a chem paper and this question always gets me. pls can you explain to me how it works i have a vague idea but it doesn't always work

8.0dm3 of NO is mixed with 6.0dm3 of O2 at room temperature and pressure (RTP).
The reaction below takes place until one of the reactants is used up.
2NO(g) + O2(g) → 2NO2(g)
What is the volume of the mixture at RTP after the reaction has taken place?
A 8.0dm3
B 10.0dm3
C 12.0dm3
D 14.0dm3
Equal volumes of gases contain the same amount of those gases. 1 dm3 of O2 contains the same number of molecules as 1 dm3 of CO2. 8 dm3 of N2 contains the same as 8 dm3 of NH3 etc.

The reaction: 2NO(g) + O2(g) → 2NO2(g) can be re-written as 2 volumes of NO react with 1 volume of O2 forming 2 volumes of NO2.

8 dm3 of NO reacts with 4 dm3 of O2 forming 8 dm3 of NO2. Note that 2 of the 6 dm3 of O2 therefore can't react and are still present at the end.

10 dm3 in total, 8 dm3 of NO2 and the 2 dm3 of O2.
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!AXEL!
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(Original post by Pigster)
Equal volumes of gases contain the same amount of those gases. 1 dm3 of O2 contains the same number of molecules as 1 dm3 of CO2. 8 dm3 of N2 contains the same as 8 dm3 of NH3 etc.

The reaction: 2NO(g) + O2(g) → 2NO2(g) can be re-written as 2 volumes of NO react with 1 volume of O2 forming 2 volumes of NO2.

8 dm3 of NO reacts with 4 dm3 of O2 forming 8 dm3 of NO2. Note that 2 of the 6 dm3 of O2 therefore can't react and are still present at the end.

10 dm3 in total, 8 dm3 of NO2 and the 2 dm3 of O2.
thank a lot for explaining it so well, it is very helpful
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amani24
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(Original post by Pigster)
Equal volumes of gases contain the same amount of those gases. 1 dm3 of O2 contains the same number of molecules as 1 dm3 of CO2. 8 dm3 of N2 contains the same as 8 dm3 of NH3 etc.

The reaction: 2NO(g) + O2(g) → 2NO2(g) can be re-written as 2 volumes of NO react with 1 volume of O2 forming 2 volumes of NO2.

8 dm3 of NO reacts with 4 dm3 of O2 forming 8 dm3 of NO2. Note that 2 of the 6 dm3 of O2 therefore can't react and are still present at the end.

10 dm3 in total, 8 dm3 of NO2 and the 2 dm3 of O2.
why is the answer not 12dm3?

the moles of NO are 0.34 and the moles of O2 are 0.25, so the O2 is limiting. I times 0.25 by 2 because of the stoichiometry and got 0.5, then x this by 24 and got 12dm3. I just don’t understand sorry! thank you
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kristenVigil
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(Original post by amani24)
why is the answer not 12dm3?

the moles of NO are 0.34 and the moles of O2 are 0.25, so the O2 is limiting. I times 0.25 by 2 because of the stoichiometry and got 0.5, then x this by 24 and got 12dm3. I just don’t understand sorry! thank you
stoichiometric ratio of NO:O2 is 2:1
basically this mean 2 moles of NO reacts with 1 mole of O2
so the limiting reagent is actually NO because 0.34/2 < 0.25
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Pigster
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(Original post by amani24)
why is the answer not 12dm3?

the moles of NO are 0.34 and the moles of O2 are 0.25, so the O2 is limiting. I times 0.25 by 2 because of the stoichiometry and got 0.5, then x this by 24 and got 12dm3. I just don’t understand sorry! thank you
0.25 mol of O2 would react with 0.5 mol of NO. BUT you only have 0.34 mol of NO, so all of the O2 can't react.
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amani24
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(Original post by kristenVigil)
stoichiometric ratio of NO:O2 is 2:1
basically this mean 2 moles of NO reacts with 1 mole of O2
so the limiting reagent is actually NO because 0.34/2 < 0.25
ahhh thank you so much. i finally understand this now !!
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amani24
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(Original post by Pigster)
0.25 mol of O2 would react with 0.5 mol of NO. BUT you only have 0.34 mol of NO, so all of the O2 can't react.
thank you!
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