H2 Chemistry - Atoms, Molecules and Stoichiometry
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hey! I admit i'm noob! haha, am a sub sci student! can someone explain in greater detail please?
1. when hydrocarbons are burned in a limited amount of air both CO and CO2 form. When 0.450g of a particular hydrocarbon was burned in air, 0.467g of CO, 0.733g of CO2 and 0.450g of H20 formed.
what is the empirical formula of the compound? (C2H3)
how many grams of O2 were used in the reaction? (1.200g)
how many grams would have been required for complete combustion? (1.467g)
2. 100cm3 of hydrogen were exploded with 160cm3 oxygen. what is the final volume if all the volumes were measured at (A) 110degcelsius and 1 atmosphere and (B) 25degcelsius and 1 atmosphere? (210cm3, 110cm3)
3. when 20cm3 of a gaseous hydrocarbonm was exploded with 150cm3 of oxygen, the residual gases(whats this?) occupied 130cm3. after shaking the products with excess sodium hydroxide the final volume is 90cm3. Deduce the molecular formula of the hydrocarbon.(all measured at rtp) (C2H4)
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Reminder :
As with all 'A' level Chemistry discussions, the following should be the chronological order :
1) Let current 'A' level forum members attempt the question posted and share their work here.
2) Let post-'A' level forum members guide the discussion.
3) I'll moderate by adding comments where necessary or helpful.
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1. Equation is CxHY + (3x+y)/4O2 ---> (x/2)CO2 + (x/2)CO + (y/2)H2O
No. of mol of CO = 0.467/28= 0.01668
No. of mol of CO2 = 0.733/44= 0.01666
No. of mol of H2O = 0.450/18=0.02500 mol
Taking whole number ratios of CO, CO2 and H2O,
no of mol of CO = 1, no of mol of CO2 =1 , no of mol of H2O = 0.025/0.01666= 1.5
So number of C in CxHy =2
number of H in CxHy = 2x 1.5 = 3 ( 2 H in H2O)
So empirical formula is C2H3
No of mol of O2 = no of mol of (2 O) in CO, CO2 and H2O = 0.01668 + 0.01666/2 + 0.025/2=0.03751 mol
Mass of O2 = 0.03751 x 32 = 1.200 g
2C2H3 + 11/2O2 ----> 4CO2 + 3H2O ( only CO2 and H2O for complete combustion)
1 mol of hydrocarbon require 11/4 mol of O2
No of mol of C2H3 = 0.450/27 = 0.01667
No of mol of O2 required = 0.01667 x 11/4 = 0.04584 mol
= 0.04584 x 32 = 1.467 g
2. 2H2 + O2 -----> 2H2O
So 2 vol of H2 react with 1 vol of O2
Therefore O2 is in excess.
(A) Water is in liquid form. Vol of O2 counted only.
Vol of O2 = 160 - 100/2 = 110 cm3
(B) Water is in gaseous form. Vol of water = 100 cm3 from equation
Total vol = water + O2 = 100 + 110 = 210 cm3
3. Residual gases refer to gases formed after reaction, including any unreacted reactants.
This include CO2 or hydrcarbon/O2 (depending on ratio)
No H2O since it is measured at rtp.
Gas absorbed by NaOH is CO2
Vol of CO2 = 130 - 90 = 40 cm3
Equation: CxHy + (x+y/4)O2 ----> xCO2 + (y/2)H2O
so x = 40/20 = 2
Vol of remaining gases minus CO2 = 90 cm3
So O2 is in excess ( not sure how to explain this)
Vol of O2 = 90 cm3 ( no H2O involved)
Then vol of O2 reacted = 150 - 90 = 60 cm3
Equation: C2Hy + (2+y/4)O2 = 2CO2 + (y/2)H2O
2+ y/4 =60/20 = 3
y = 4
So molecular formula of hydrocarbon is C2H4
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Ok... shall refrain from posting first... I post A lvl lol
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Originally posted by Chemfreak022:
Ok... shall refrain from posting first... I post A lvl lol
No need to refrain at all, Chemfreak022, because you realize that many current 'A' level forum members are hesistant to post and share, because being current 'A' level students, they're not as confident as you guys (post-A-levellers) are.So feel free to post, otherwise, especially for the more difficult questions, it may be days before any current 'A' level students dare to attempt the question. And the question poster (TS or OP; ThreadStarter or OriginalPoster) may become impatient, or may need to submit the homework assignment soon.
So everyone, current 'A' level students as well as post-A-level pple, feel free to post.
I'll only add comments when necessary. Moderators aren't homework workers, or thread-hoggers, we're only here to keep discussions in check, and to add to discussion where helpful. Besides and furthermore, I noticed more and more post-A-level pple are visiting the homework forum to help out (obviously it can't be "to learn from", since they're / you guys are already beyond 'O' and 'A' levels). So it's good if the forum gets more autonomous in activity, without needing us moderators to spoonfeed answers in every thread.
Keep up the helpful contributions, Chemfreak022, and all other post-A-Levellers!
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Originally posted by Chemfreak022:
1. Equation is CxHY + (3x+y)/4O2 ---> (x/2)CO2 + (x/2)CO + (y/2)H2O
No. of mol of CO = 0.467/28= 0.01668
No. of mol of CO2 = 0.733/44= 0.01666
No. of mol of H2O = 0.450/18=0.02500 mol
Taking whole number ratios of CO, CO2 and H2O,
no of mol of CO = 1, no of mol of CO2 =1 , no of mol of H2O = 0.025/0.01666= 1.5
So number of C in CxHy =2
number of H in CxHy = 2x 1.5 = 3 ( 2 H in H2O)
So empirical formula is C2H3
No of mol of O2 = no of mol of (2 O) in CO, CO2 and H2O = 0.01668 + 0.01666/2 + 0.025/2=0.03751 mol
Mass of O2 = 0.03751 x 32 = 1.200 g
2C2H3 + 11/2O2 ----> 4CO2 + 3H2O ( only CO2 and H2O for complete combustion)
1 mol of hydrocarbon require 11/4 mol of O2
No of mol of C2H3 = 0.450/27 = 0.01667
No of mol of O2 required = 0.01667 x 11/4 = 0.04584 mol
= 0.04584 x 32 = 1.467 g
2. 2H2 + O2 -----> 2H2O
So 2 vol of H2 react with 1 vol of O2
Therefore O2 is in excess.
(A) Water is in liquid form. Vol of O2 counted only.
Vol of O2 = 160 - 100/2 = 110 cm3
(B) Water is in gaseous form. Vol of water = 100 cm3 from equation
Total vol = water + O2 = 100 + 110 = 210 cm3
3. Residual gases refer to gases formed after reaction, including any unreacted reactants.
This include CO2 or hydrcarbon/O2 (depending on ratio)
No H2O since it is measured at rtp.
Gas absorbed by NaOH is CO2
Vol of CO2 = 130 - 90 = 40 cm3
Equation: CxHy + (x+y/4)O2 ----> xCO2 + (y/2)H2O
so x = 40/20 = 2
Vol of remaining gases minus CO2 = 90 cm3
So O2 is in excess ( not sure how to explain this)
Vol of O2 = 90 cm3 ( no H2O involved)
Then vol of O2 reacted = 150 - 90 = 60 cm3
Equation: C2Hy + (2+y/4)O2 = 2CO2 + (y/2)H2O
2+ y/4 =60/20 = 3
y = 4
So molecular formula of hydrocarbon is C2H4
Wah, fierce. You got so much free time in your hand? Really step by step sia.
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Originally posted by d3sT1nY:
Wah, fierce. You got so much free time in your hand? Really step by step sia.
Pple helpful that's why so detailed, don't say "so much free time ar?". Let's be appreciative k? -
Originally posted by UltimaOnline:
Pple helpful that's why so detailed, don't say "so much free time ar?". Let's be appreciative k?Haha, just some comments. No hard feeling. =P
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Originally posted by UltimaOnline:
I'll only add comments when necessary. Moderators aren't homework workers, or thread-hoggers, we're only here to keep discussions in check, and to add to discussion where helpful. Besides and furthermore, I noticed more and more post-A-level pple are visiting the homework forum to help out (obviously it can't be "to learn from", since they're / you guys are already beyond 'O' and 'A' levels). So it's good if the forum gets more autonomous in activity, without needing us moderators to spoonfeed answers in every thread.
Keep up the helpful contributions, Chemfreak022, and all other post-A-Levellers!
Besides and furthermore, I noticed more and more post-A-level pple are visiting the homework forum to help out (obviously it can't be "to learn from", since they're / you guys are already beyond 'O' and 'A' levels).
Our initial hardwork paid off, and more pple are visiting to help out after they have gained from the forum (hopefully)
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Originally posted by eagle:
Our initial hardwork paid off, and more pple are visiting to help out after they have gained from the forum (hopefully)
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Originally posted by Chemfreak022:
2. 2H2 + O2 -----> 2H2O
So 2 vol of H2 react with 1 vol of O2
Therefore O2 is in excess. (shouldn't H2 be in excess since 2 vol of H2 react with 1 vol of O2)
(A) Water is in liquid form. Vol of O2 counted only.
Vol of O2 = 160 - 100/2 = 110 cm3
(B) Water is in gaseous form. Vol of water = 100 cm3 from equation
Total vol = water + O2 = 100 + 110 = 210 cm3
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Hence hydrogen is the limiting reactant and oxygen is the excess reactant.
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hey. can anyone gimme a quick answer to this question, cause i'm stuck.
(COOH)2 + NaOH -> what's the salt? + H20
i can balance it on my own! haha. thanks
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COONa. This is O-level stuff btw.
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sry! i'm combined sci.. i'm chocolates-xed aniwae
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Originally posted by blueftw:
sry! i'm combined sci.. i'm chocolates-xed aniwae
Why clone?
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hey can i ask a question? how to explain the difference between 0.9 moldm-3 aqueous hcl and 0.900 molddm-3 aqueous hcl.
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Can some1 teach me how to balance redox using half equation method?
EG: Cu + HNO3 -> Cu(NO3)2 + NO
I2 + NA2S2O3 -> NAI + NA2S4O6 -
Originally posted by Destined2REIGN:
Can some1 teach me how to balance redox using half equation method?
EG: Cu + HNO3 -> Cu(NO3)2 + NO
I2 + NA2S2O3 -> NAI + NA2S4O6Here is how you balance half-equations :
Hydrogen peroxide is reduced to water (chemical name : hydrogen hydroxide aka dihydrogen monoxide aka hydrogen(I) oxide).
Hydrogen peroxide is oxidized to molecular oxygen gas.
(Acidified) Manganate(VII) ions are reduced to Manganese(II) cations.
Nitrogen dioxide is oxidized to Nitrate(V) anions.
[reduction]
H2O2 --> H2O
To balance oxygen,
H2O2 --> 2H2O
To balance hydrogen,
2H+ + H2O2 --> 2H2O
To balance charges,
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O2 --> O2
To balance hydrogen,
H2O2 --> O2 + 2H+
To balance charges,
H2O2 --> O2 + 2H+ + 2e-
[reduction]
MnO4 - --> Mn2+
To balance oxygen,
MnO4 - --> Mn2+ + 4H2O
To balance hydrogen,
8H+ + MnO4 - --> Mn2+ + 4H2O
To balance charges,
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
[oxidation]
NO2 --> NO3 -
To balance oxygen,
H2O + NO2 --> NO3 -
To balance hydrogen,
H2O + NO2 --> NO3 - + 2H+
To balance charges,
H2O + NO2 --> NO3 - + 2H+ + e-
To obtain overall balanced redox equations, we ensure no. of electrons lost = no. of electrons gained (by multiplying one or both half-equations to obtain the lowest common multiple for coefficient of electrons), then add up the half equations (left hand side + left hand side, right hand side + right hand side), and finally cancel away any common species on both sides, invariably including all electrons.
For the reaction of using acidified KMnO4(aq) on H2O2, we have
[reduction]
8H+ + MnO4 - + 5e- --> Mn2+ + 4H2O
16H+ + 2MnO4 - + 10e- --> 2Mn2+ + 8H2O
[oxidation]
H2O2 --> O2 + 2H+ + 2e-
5H2O2 --> 5O2 + 10H+ + 10e-
[Balanced Redox]
(16H+ + 2MnO4 - + 10e-) + (5H2O2) --> (2Mn2+ + 8H2O) + (5O2 + 10H+ + 10e-)
6H+ + 2MnO4 - + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
For the reaction of bubbling NO2 into H2O2, we have
[reduction]
2H+ + 2e- + H2O2 --> 2H2O
[oxidation]
H2O + NO2 --> NO3 - + 2H+ + e-
2H2O + 2NO2 --> 2NO3 - + 4H+ + 2e-
[Balanced Redox]
(2H2O + 2NO2) + (2H+ + 2e- + H2O2) --> (2NO3 - + 4H+ + 2e-) + (2H2O)
2NO2 + H2O2 --> 2NO3 - + 2H+
The Question :
300 cm3 of a mixture of dinitrogen monoxide and nitrogen dioxide, at r.t.p. conditions is bubbled through 75 cm3 of 0.10 mol/dm3 acidified hydrogen peroxide solution. The nitrogen dioxide is oxidized to nitrate(V) ions, while the inert dinitrogen monoxide does not react. The remaining hydrogen peroxide in 50.0 cm3 of the resulting solution is then titrated with a 0.050 mol/dm3 acidified potassium manganate(VII) solution, of which 22.0 cm3 was required. Calculate the % by volume of dinitrogen monoxide in the gas mixture.
Final Answer :
46%.
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Here's a useful list of redox half equations.
Note that you can obtain the oxidation potential by changing the -ve/+ve sign. For instance,
The standard reduction potential for Ag+ to Ag is +0.8V.
The standard oxidation potential for Ag to Ag+ is -0.8V.
The standard reduction potential for Zn2+ to Zn is -0.76V.
The standard oxidation potential for Zn to Zn2+ is +0.76V.
