anonymous
  • anonymous
http://oi68.tinypic.com/sfv70p.jpg
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jebonna
  • jebonna
Well first off looking at the diagram we can see the reactants and the products. From this we need to make an equation. In the diagram it is helpful that each element has a key. From this we can work out how many of each element/compound there are. In the first box we see: 2 lots of NH3 and also two lots of CIF3. In the second box we see: 1 lot of N2, 1 lot of Cl2 and 6 lots of HF. Now we can make an equation: 2 NH3 + 2 ClF3 = N2 + Cl2 + 6 HF Lets do the rest in steps: 1. Now to work out grams of an element we first need to work out the moles. The equation for this is: mass/molar mass You can use either one of the masses (g) you have and molar mass is the mass of NH3 or CIF3 (add all the relative atomic masses together from periodic table) 2. Once you have done this, you will have the moles for either NH3 or CIF3 (which ever one you chose). You then need to deduce the moles (work out the moles for the products you're trying to find the mass for). Because the moles of whatever you worked out is x2 (because in the equation there are 2 lots of NH3 and CIF3), you will need to half the answer you got by 2 to get the moles for N2 and Cl2 (because there is only 1 lot of both of those in the equation). Then, to get the moles for HF, you will need to times the answer by 3, because there are 6 lots of HF's in the equation, and the answer you have is for 2 moles. 3. Once you have done that for each product (N2, Cl2 and HF), you can then work out the mass in grams for all of them. To do this you use this equation: molar mass x moles You take the moles of each product (that you worked out earlier) and times it, by each products molar mass. For example, the molar mass for HF would be hydrogen's relative atomic mass (on periodic table) add fluorine's relative atomic mass. Do that for each product, and then you will end up with the answer in grams, for each one. 4. To check you were write, you can add the grams of the reactants at the beginning together to get a total number, and then in a separate calculation, add the mass in grams of all of the products together. The numbers should be the same (due to the law of conservation of mass). I hope this can help, and I hope you can work it out! If you have any problems, feel free to message me. :)
anonymous
  • anonymous
@matt101
matt101
  • matt101
I agree with exactly what @jebonna said. Follow those steps carefully and you'll get the answer!

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anonymous
  • anonymous
I'm getting confused with what they are saying though.
anonymous
  • anonymous
@matt101
matt101
  • matt101
What don't you understand?
anonymous
  • anonymous
What exactly I'm need to set up and what goes where. I need to see that visually I think.
anonymous
  • anonymous
I tried to set up a problem a little bit
anonymous
  • anonymous
|dw:1446925608633:dw|
anonymous
  • anonymous
|dw:1446925892849:dw|
anonymous
  • anonymous
Is that done right?
matt101
  • matt101
Ok let's start from the beginning. Here's the basic outline of what needs to happen (paraphrasing what was written above) - let me know if this isn't clear: STEP 1: Come up with a chemical reaction. This will let us figure out how many products we should expect based on the reactants used. STEP 2: Find moles of each reactant (remember, we can't figure out amounts using masses), and determine the limiting reactant (since we're given masses of both starting reactants). STEP 3: Calculate the moles of each product you should get based using stoichiometry and the chemical reaction you came up with in step 1. STEP 4: Convert the moles of each product back to masses. Start with step 1. What is your reaction?
anonymous
  • anonymous
2 NH3 + 2 ClF3 = N2 + Cl2 + 6 HF
anonymous
  • anonymous
Was what I put above in the drawing remotely right?
matt101
  • matt101
Sorry you posted just as I did! Let me have a look
anonymous
  • anonymous
It's fine.
matt101
  • matt101
Ok what you did is close, but you're going to want to divide by 2 (the stoichiometric coefficient) rather than multiply by 2 to determine which is the limiting reactant. However, the number you get before dividing by 2 are the ACTUAL moles of each reactant you're working with, and these are the numbers you will use to calculate the moles of each product.
anonymous
  • anonymous
|dw:1446926447363:dw|
anonymous
  • anonymous
So would it be like that instead. Obviously with a different answer.
matt101
  • matt101
You still need to divide by 2. I get that you divided what you HAD by 2, but what I meant was that you need to divide originally INSTEAD of multiply. And yes with different answers.
anonymous
  • anonymous
Is the 2 coming from the chemical equation?
matt101
  • matt101
Yes!
anonymous
  • anonymous
|dw:1446926721648:dw|
matt101
  • matt101
Right and after you do the same for the other reactant, which is the limiting reactant?
anonymous
  • anonymous
|dw:1446927012542:dw|
anonymous
  • anonymous
It would be the first one.
matt101
  • matt101
Great so now all stoich comparisons will be need to be made to NH3. You've already found the moles of each reactant (the original numbers, before dividing by 2), so use the appropriate stoich ratios to find the moles of each product.
anonymous
  • anonymous
|dw:1446927529338:dw|
matt101
  • matt101
Right, so those are moles of reactants - NH3 and ClF3 - that actually react. Now use that information to find the moles of each product
anonymous
  • anonymous
I'm confused.
matt101
  • matt101
Focus on NH3, since that's the limiting reactant. We have 1.6 mol of NH3 (27.3 g / 17.031 g/mol). We want to find moles of each product. Use stoich ratios to figure this out. As an example, for N2: x mol N2 / 1.6 mol NH3 = 1 N2 / 2 NH3 The left side are the actual moles of stuff you're working with in the reaction, the right side are the stoich coefficients taken from the balanced chemical reaction. Solve for x and you get 0.8 mol of N2 produced. Now do the same for Cl2 and HF!
anonymous
  • anonymous
|dw:1446929131805:dw|
anonymous
  • anonymous
At first would it look like this?
matt101
  • matt101
No - please reread my answer carefully!
anonymous
  • anonymous
|dw:1446929302563:dw|
matt101
  • matt101
Again, look carefully at my answer above. You need to solve for x, except now it'll be fore Cl2 rather than N2. What is x in that case?
anonymous
  • anonymous
I'm not following along. Sorry.
matt101
  • matt101
x mol Cl2 / 1.6 mol NH3 = 1 Cl2 / 2 NH3 Solve for x!
anonymous
  • anonymous
Okay so do I need to cross multiply?
matt101
  • matt101
Yes. You can get rid of all the words if that's easier, I've just left them in so you can see what each number is referring to: x/1.6 = 1/2 What do you get for x?
anonymous
  • anonymous
2x = 1.6
anonymous
  • anonymous
x=.8
matt101
  • matt101
Great! Now do the same thing for HF. Be sure to use the right coefficients in your equation.
anonymous
  • anonymous
|dw:1446931145587:dw|
anonymous
  • anonymous
Like that?
matt101
  • matt101
EXACTLY. What do you get?
anonymous
  • anonymous
x =4.8
anonymous
  • anonymous
@matt101 is that right?
matt101
  • matt101
Yup! Now that you've found moles of each product, multiply each by the respective molar mass, and the three masses you get are your answers!
anonymous
  • anonymous
Do I do the 3 masses separately?
matt101
  • matt101
Yes. For each product, multiply the moles of product you calculated by the molar mass of that product.
anonymous
  • anonymous
|dw:1446932711745:dw|
anonymous
  • anonymous
|dw:1446932988843:dw|
matt101
  • matt101
Yup keep going
anonymous
  • anonymous
|dw:1446933154279:dw|
matt101
  • matt101
Yes and the last one?
anonymous
  • anonymous
Was the 1.6 mol for the N2?
anonymous
  • anonymous
Never mind what was for NH3. What was the mol for N2?
anonymous
  • anonymous
.8013 mol was N2 right?
matt101
  • matt101
Yes
matt101
  • matt101
Anyways once you have that third mass, you've answered the question. Please read through what we did above carefully and make sure you understand all the steps so that you'll be able to do this kind of thing next time no problem!
anonymous
  • anonymous
|dw:1446933994587:dw|
anonymous
  • anonymous
@matt101 thanks for the help!

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