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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: inorganiic on March 25, 2021, 04:21:21 AM

Title: mixture
Post by: inorganiic on March 25, 2021, 04:21:21 AM
The mixture of hydrogen, methane and co gases under normal conditions has a density of 0.857g per dm3. 4.52 volumes of air are required for complete combustion of one volume of the mixture. Determine the composition of the mixture in percent. Can someone hel pe
Title: Re: mixture
Post by: sjb on March 25, 2021, 05:05:08 AM
Where have you got stuck? Can you, for instance, write individual equations for each of the reactions going on?
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 05:23:02 AM
OK
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 05:30:30 AM
Oh am i suposed to use 21 percent o2 in air?
CH4 + O2 ---> CO2 + 2H2O
2CO + O2 --->2CO2
2H2 + O2---> 2H2O
Title: Re: mixture
Post by: Borek on March 25, 2021, 05:42:04 AM
Quote from: inorganiic on March 25, 2021, 05:30:30 AM
Oh am i suposed to use 21 percent o2 in air?

Yes.

Hint: you have to use information given to build a system of equations where your unknowns are percentages of gases in the mixture.
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 05:57:41 AM
Can i calculate V of O2 and make proportion for each reaction but actually i don`t know how to calculate this and what to use for 0.857 g per dm3
Title: mixture
Post by: inorganiic on March 25, 2021, 08:11:19 AM
sorry for posting here
The mixture of hydrogen, methane and co gases under normal conditions has a density of 0.857g per dm3. 4.52 volumes of air are required for complete combustion of one volume of the mixture. Determine the composition of the mixture in percent.
CH4 + O2 ---> CO2 + 2H2O
2CO + O2 --->2CO2
2H2 + O2---> 2H2O
can someone solve this, thank you so much
Title: Re: mixture
Post by: chenbeier on March 25, 2021, 08:20:04 AM
Burning of methane is wrong.
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 08:24:15 AM
I know 2 before O2 but how to solve this
Title: mixture
Post by: inorganiic on March 25, 2021, 09:49:51 AM
sorry for posting here
The mixture of hydrogen, methane and co gases under normal conditions has a density of 0.857g per dm3. 4.52 volumes of air are required for complete combustion of one volume of the mixture. Determine the composition of the mixture in percent.
CH4 + 2O2 ---> CO2 + 2H2O
2CO + O2 --->2CO2
2H2 + O2---> 2H2O
can someone solve this, thank you so much
Title: Re: mixture
Post by: billnotgatez on March 25, 2021, 10:24:04 AM
@ inorganiic
We only post a given question once on the forum.
This is a forum policy.
Click on the link near the top center of the forum page.
Forum Rules: Read This Before Posting.
http://www.chemicalforums.com/index.php?topic=65859.0

I moved all your posts to here so as not to have duplicates
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 12:04:15 PM
oh i didn't knew and i'm so sorry. Can you solve this problem or tell me how
Title: Re: mixture
Post by: Borek on March 25, 2021, 03:04:16 PM
Assume fraction of gases to be x, y, z. If these are fractions - what is their sum?

Assume there is 1L of the mixture - can you express its mass using x, y, z? how is it related to the mixture density?
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 03:15:03 PM
ok i calculated that the mass of the mixture is 0.857g but how to set up the system and what is the volume of air used for
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 03:47:05 PM
can I solve without the system in some way and how to use the data of 4.52 air volume. does that mean 4.52 * 0.21 for each equation
Title: Re: mixture
Post by: Borek on March 25, 2021, 05:44:23 PM
Quote from: inorganiic on March 25, 2021, 03:15:03 PM
ok i calculated that the mass of the mixture is 0.857g

That was given, so you have not calculated it - but perhaps you expressed this information using an equation. Hard to say not seeing your work.

Quote
but how to set up the system and what is the volume of air used for

Volume and mass are required to express the density.

Quote from: inorganiic on March 25, 2021, 03:47:05 PM
can I solve without the system in some way

No.

Quote
and how to use the data of 4.52 air volume. does that mean 4.52 * 0.21 for each equation

4.52*0.21 is the volume of oxygen required to burn all gases.

Assuming you have n liters of CH4, what is the volume of oxygen required to burn it?
Title: Re: mixture
Post by: Orcio_87 on March 25, 2021, 05:44:41 PM
@inorganiic

1. Calculate density of CH4, H2 and CO at normal conditions (T = 20 C, p = 1 atm).
2. If mixture composes of 100 % CH4 it will have density you have calculated. If 100 % H2 - some other density. If 100 % CO - yet another density.
3. Calculate how many litres of air is needed to burn 1 litre of pure CH4, 1 litre of pure H2, and 1 litre of pure CO.
4. You know that mixture contains only CH4, H2 and CO.
5. Points 2, 3 and 4 gives you system of equations (3 equations with 3 variables) to solve.
Title: Re: mixture
Post by: inorganiic on March 25, 2021, 06:15:05 PM
@Orcio_Dojek
I still can't solve it, I can't do the system, I calculated the densities 1.1515, 0.3290, 0.1645 and how much it takes to burn 1 dm3, 0,526 for CO and H2 and 2,107 but it doesn't work. Can you solve it I need for the entrance test. Solution is 20 percent for H2, 30 percent for CH4 and 50 percent for CO
Title: Re: mixture
Post by: Orcio_87 on March 26, 2021, 06:20:49 PM
Quote
I calculated the densities 1.1515, 0.3290, 0.1645

What the hell? These numbers can't be correct.

Also:
Quote
how much air it takes to burn 1 dm3, 0,526 for CO and H2 and 2,107 for CH4

I wonder how it is possible if mixture cointains only CH4, H2 and CO, and for one liter it is required 4,52 liters of air ???
Title: Re: mixture
Post by: inorganiic on March 27, 2021, 03:33:55 PM
Can you solve it  because i can not almoast three days
Title: Re: mixture
Post by: Orcio_87 on March 27, 2021, 04:05:22 PM
1. Density of CH4, CO, H2 and density of their mixture.
2. How much air is needed to burn 1 liter of CH4, CO, H2 and their mixture.
3. Mixture cointains only CH4, CO and H2.

Rest is only calculation.
Title: Re: mixture
Post by: inorganiic on March 27, 2021, 05:07:25 PM
But that is a problem i can not get 20 30 and 50 percent
Title: Re: mixture
Post by: Orcio_87 on March 27, 2021, 05:15:04 PM
I know. Concentration of H2 is lower than given answer.

Maybe author mistaken "normal" with "standard" (T = 273 K) conditions, I don't know.
Title: Re: mixture
Post by: inorganiic on March 27, 2021, 06:15:08 PM
Can you text this here how you got the answer, i would be so thankful
Title: Re: mixture
Post by: Orcio_87 on March 27, 2021, 06:43:04 PM
@inorganiic

I must see your calculations before that.
Title: Re: mixture
Post by: inorganiic on March 27, 2021, 06:53:29 PM
In normal conditions we need 2,38l of air for H2 and CO from proportion, also when i made proportion i get 0,5l of O2 then for CH4 2l of O2 and 9,521l of air and then all is complicated and i can not get the right solution
Title: Re: mixture
Post by: Orcio_87 on March 28, 2021, 04:43:47 AM
I know. My results (concentration of H2) also differ slightly from that given by the author.
Title: Re: mixture
Post by: inorganiic on March 28, 2021, 05:01:07 AM
Can you write me how you got your results detailed
Title: Re: mixture
Post by: inorganiic on March 28, 2021, 05:05:32 AM
Or better photo them if you can
Title: Re: mixture
Post by: Borek on March 28, 2021, 10:39:54 AM
Quote from: inorganiic on March 28, 2021, 05:01:07 AM
Can you write me how you got your results detailed

Quote from: inorganiic on March 28, 2021, 05:05:32 AM
Or better photo them if you can

Per the forum rules (https://www.chemicalforums.com/index.php?topic=65859.0) you agreed to when registering we don't give solutions. Begging and pestering won't change that.

However, if you post how you approached the problem (and you were given multiple hints) we will gladly help you solve the problem.