Chemistry

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Use the following information to identify element A and come
pound B, then answer questions a and b.

An empty glass container has a mass of 658.572 g. It has a
mass of 659.452 g after it has been filled with nitrogen gas at a
pressure of 790. torr and a temperature of 15°C. When the container
is evacuated and refilled with a certain element (A) at a
pressure of 745 torr and a temperature of 26"C, it has a mass of
660.59 g.

Compound B, a gaseous organic compound that consists of
85.6% carbon and 14.4% hydrogen by mass, is placed in a stainless
steel vessel (10.68 L) with excess oxygen gas. The vessel is
placed in a constant-temperature bath at 22°C. The pressure in
the vessel is 1 1.98 atm. In the bottom of the vessel is a container
that is packed with Ascarite and a desiccant. Ascarite is asbestos
impregnated with sodium hydroxide; it quantitatively absorbs
carbon dioxide:

2NaOH(s) + CO2(g) -----> Na2CO3(s) H2O(l)

The desiccant is anhydrous magnesium perchlorate, which
quantitatively absorbs the water produced by the combustion reaction
as well as the water produced by the above reaction. Neither
the Ascarite nor the desiccant reacts with compound B or
oxygen. The total mass of the container with the Ascarite and
desiccant is 765.3 g.


The combustion reaction of compound B is initiated by a
spark. The pressure immediately rises, then begins to decrease,
and finally reaches a steady value of 6.02 atm. The stainless steel
vessel is carefdly opened, and the mass of the container inside
the vessel is found to be 846.7 g.



A and B react quantitatively in a 1: 1 mole ratio to form one
mole of the single product, gas C.
a. How many grams of C will be produced if 10.0 L of A and
8.60 L of B (each at STP) are reacted by opening a stopcock
connecting the two samples?
b. What will be the total pressure in the system?

  • Chemistry -

    You need to tell us what you have tried. Surely you know where to start. Can you identify element A from the early information? Then what don't you understand about the remainder?

  • Chemistry -

    Well I got Chlorine gas for Element, I just don't understand how to find Compound B and answer the questions.

  • Chemistry -

    Very good. Cl2 is element A. The empirical formula for compound B can be determined from the percent composition, which is given. Let me know what you find for the empirical formula of compound B.

  • Chemistry -

    The empirical formula for compound B is CH2

  • Chemistry -

    OK. Your are correct. The empirical formula is CH2.
    Use PV = nRT to solve for mols gas at 11.98 atm pressure (before combustion) and use PV = nRT to solve for mols gas at 6.02 (after combustion), then subtract to determine the mols of compound B plus oxygen. Alternatively, you can subtract 11.98 - 6.02 = 5.96 atm and use PV = nRT to solve for the mols of compound B + oxygen that reacted.
    From the container that had a mass of 846.7 g after combustion and 765.3 g before combustion, subtraction will give the mass of the combined CO2 + H2O (the ascarite and magnesium perchlorate absorb all the CO2 and all the water). You need to know how to divide that 81.4 g into the amount for CO2 and the amount for H2O. To do that, it is easy to look at the equation for a few of the simple gases with CH2 as the empirical formula. C2H4, C3H6, and C4H8 are the first three. Write equations for the combustion of all three of those and you will see that all of them produce equal mols of CO2 and H2O. The mol ratios of the reactants are different but the products are always equal. SO, divide up the CO2 and H2O produced in the combustion so you have equal mols of CO2 and H2O. You can do that by letting X = mass CO2 and 81.4-X = mass H2O.
    Then X/44.01 = (81.4-X)/18.015. Solve for X to get grams CO2 and grams CO2 and they should add to 81.4 g. Then determine mols CO2 and mols H2O from that (They will be equal). Using your equations, convert mols CO2 to mols compound B and determine mols O2 needed for combustion. Add them together. There is only one that meets the criteria for mols used in the reaction and that is C2H4.
    I will leave the questions a and b to you but let me know if you have any problems with them. Provide your work so I can better help if you do need additional assistance on a or b.

  • Chemistry -

    I get everything, but what's the use of finding the moles when u subtract 6.02 from 11.98, what do you do with the moles?

    And for Part A im getting 37grams
    Part B I got .83 atm..

    Am I right?

  • Chemistry -

    5.96 mol compound B plus O2 reacted and n = 2.63 B + O2. From my last response, I found 81.4 g combined CO2 + H2O = 1.31 mols CO2 and 1.31 mols H2O.
    C2H4 + 3O2 ==> 2CO2 + 2H2O
    So 1.31/2 mol C2H4 = 0.6567 mols CH4.
    3 x 0.6567 = mols O2 = 1.965
    0.6567 + 1.965 = 2.62 which agrees with the total mols from the 5.96 pressure reduction when the compound was combusted. You can then convert the combined CO2 + H2O from the 81.4 to g C and g H, combine them (I get 15.76 g C and 2.625 g H = 18.385 g) and that = 0.6567 mol CxHy. Then since mols = g/molar mass, rearranging get molar mass = g/mols = 018.385/0.6567 = 27.996 which rounds to 28 and that tells you the compound is (CH2)2 = C2H4 and that has a molar mass of 28.
    I'll look at your answers for part a and b later. Don't have time now.

  • Chemistry -

    You didn't show your work so I can't find your error but I obtained 37.97 g for part a and that rounds to 38 g. For part b I found 0.537 atm which rounds to 0.54 atm. Check my arithmetic.

  • Chemistry -

    I think we're both right about the 37.9 grams or whatever, and I think you're right about the .538 atm, because one of the answer keys says so, but how did you get .538 atm?

    By the way, I appreciate your help on this problem, thanks so much, I owe u.

  • Chemistry -

    I think I did it the hard way but I converted 10 L C2H4 to mols and 8.60 L Cl2
    to mols.
    mols C2H4 (A) = 10 L/22.4 = 0.446
    mols Cl2 (B)= 8.6/22.4 = 0.384
    mols product formed (C)= 0.384
    mols A remaining unreacted = 0.446-0.384 = 0.062.
    Total mols = 0.384 + 0.062 = 0.446
    PV = nRT
    p = nRT/V = 0.446*0.08205*273/18.6 = 0.537 atm.

  • Chemistry -

    putting this answer online hurts alot of teachers who try to use marathon problems as a learning tool

  • Chemistry -

    shut up anonymous, i love you bob

  • Chemistry -

    I don't understand how you get the mols product formed of (c) and why there are moles of a that remain unreacted

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