chemistry question
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
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
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
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
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
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.
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.
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
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.
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
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
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
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
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.
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
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 !!
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!
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.
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.
Can you please explain where the 4dm3 came from
Original post by ilayda 1234
Can you please explain where the 4dm3 came from
The molar ratio of the equation shows that 2 molecules of NO reacts with 1 molecule of O2.
So... 2 moles of NO reacts with 1 mole of O2.
2 mol is 48 dm3 and 1 mol is 24 dm3.
But these 48 and 24 volumes can be divided...
Surely 24 dm3 will react with 12 dm3 or 6 dm3 and 3 dm3. Basically, whatever volume of NO you have, you need half that volume of O2.
So, if you have 8 dm3 of NO (like it says in the Q) then you need 4 dm3 of O2.
BUT we have 6 dm3, i.e. we have 2 dm3 of O2 more than we need. So that 2 dm3 of O2 has nothing to react with (once the 8 dm3 of NO has reacted)
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