I have a few questions, if anyone can help me at all it would be amazing. i need this done by TOMORROW (03/03/2008)

1. Cyanogen has been observed in the atmopshere of Titan, Saturn's largest Moon, and in the gases of the interstellular nebulas. On Earth, it is used as a welding gas and a fumigant. In its reaction with floruine gas, carbon tetrafluroide and nitrogen trifluroride are produced. What mass of carbon tetrafluoride forms when 80.0 grams of each reactant are mixed?

2. An Element "X" has a triiodide with a formula of XI3 (the 3 should be subscript) and a trichloride with a formula of XCL3. The triioidide is quantitantively converted to the trichloride when it is heated in a stream of chlorine, according to the equation:
2XI3 + 3CI2 -> 2XCI3 + 3I2

If 0.8000 grams of XI3 ius treated, 0.3776 grams of CLI3 is obtained.

a.) Calculate the atomic mass of the element X

b.) Identify the element X

3. Hydrogen gas has been suggested to clean fuel because it produces only water vapor when it burns in the presence of O2. If the reaction has a 98.8% reaction yeild, what mass of hydrogen forms from 85.0 grams of water?

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  1. One question to a post please.

    1. Write the equation and balnace it.
    2. This is a limiting reagent problem.
    2a. Convert 80 g (CN)2 to mols. mols = grams/molar mass.
    2b. Convert 80 g F2 to mols.

    3. Using the coefficients in the balanced equation,
    3a. convert mols (CN)2 to mols CF4.
    3b. convert mols F2 to mols CF4.
    3c. Pick the smaller of the mols as the correct value. That will be the limiting reagent and the number of mols CF4.

    4. Convert mols to grams.
    mols x molar mass = grams.

    Problem 2. You need to clarify this problem. Nowhere do you show where the ClI3 comes from.

    Problem 3. This is a simple stoichiometry problem. Do it the same way as #1 except you omit the steps 2b and 3b and 3c since you know what the limiting regent is. It is 85.0 g H2O. When you finish calculating the amount of hydogen, multiply it by 0.988 to account for the 98.8% efficiency of the reaction.
    Post your work if you get stuck.

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    i actually typed number two exactly as written. maybe that's why it is so confusing to me too - it really does not contain all the infomation needed, at least i do not think it does.

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  3. When you say in step 2a to convert you mols by using grams/molar mass, you do mean to just put 80/the molar mass of (CN)2 correct?

    because i have been taught to do it differently.

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  4. Right.
    mols (CN)2 = grams cyanogen/molar mass cyanogen.
    You may have been taught not to use a formula, which is what I wrote in my first resonse, but to use dimensional analysis.
    grams (CN)2 x [1mol (CN)2/molar mass (CN)2)]= mols (CN)2.

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  5. youre right. we were taught to use dimenstional analysis as opposed to a formula.

    i'm a bit confused at this point now.

    3a. convert mols (CN)2 to mols CF4.

    I'm not sure how to set it up. Do i put the mols of (CN)2 I found (which is 2.67 by my caluculations) over 2CF4? Sorry for all the questions.

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  6. i figured out my question, haha. will post if i need more help. thanks again! this website is amazing!

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  7. Since you are so familiar with dimensional analysis, let me show you how that conversion works. But first, I don't get 2.67 mols (CN)2.
    80 g (CN)2 x [1 mol (CN)2/52.04 g (CN)2] = 1.54 mols (CN)2
    Now to convert to mols CF4.
    mols CF4 = mols (CN)2 x [2 mols CF4/1 mol (CN)2] = 1.54 mols (CN)2 x [2 mols CF4/1 mol (CN)2] = 1.54 x 2/1 = 3.07 mols CF4.
    Note we start with (CN)2. We want to get rid of the (CN)2 and end up with units of CF4. So placing the mols (CN)2 on the bottom cancels with (CN)2 we started with, and leaving the mols CF4 on top and it cancels with nothing. So we got rid of the mol (CN)2 and are left with mols CF4. It works EXACTLY the same as converting grams to mols. And if you don't know which way the factor goes; i.e., is it [1 mol (CN)2/2 mol CF4] or [2 mol CF4/1 mol (CN)2], you can mentally look at the numbers and know which will cancel and which will not.

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