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taramgrant0543664

  • one year ago

Hey everyone! I put together a basic tutorial on finding your theoretical yield as I had noticed that I had answered quite a few of these on here within the past couple weeks and I have to teach it to my brother this week so if you have any suggestions on how I can improve this that would be great!

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  1. taramgrant0543664
    • one year ago
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    I will go through a couple of questions and explain what I do as it is easier to understand with examples (in my opinion). Example 1: 3.00g of Na (sodium) react with 5.00g of Cl2 (chlorine) to form NaCl (table salt) according to the following equation: 2Na + Cl2 --> 2NaCl What is the theoretical yield of NaCl in grams? To do this we first have to determine our limiting reagent between the Na (molar mass= 22.99g/mol) and the Cl2 (molar mass=70.90g/mol) I will do the sodium first: 3.00g Na * (1mol Na/22.99g Na)=0.13mol Na 5.00g Cl2 *(1 molCl2/70.90g Cl2)=0.71mol Cl2 We also have to look at the stoichiometric coefficients of the reactants and the products the Na and NaCl have a 1:1 ratio so that means that 0.13mols of Na are used and 0.13mols of NaCl are produced but the Cl2 and the NaCl has a different ratio of 1:2 so that meant that 1.42 moles of NaCl are produced for every 0.71mols of Cl are used. Knowing this we know that the limiting reagent is the Na. Since we have determined the limiting reagent we can determine the mass of NaCl (molar mass of 58.44g/mol) are produced. 0.13mol Na * (1mol NaCl/1mol Na) * (58.44g NaCl/1mol NaCl)=7.60g of NaCl The theoretical yield of NaCl is 7.60g Example 2: You are given the following reaction: 2H2(g)+O2(g)→ 2H2O. You have 1.5mols of H2 and 5.00mols of O2 how much H2O is produced and much of the non-limiting reagent do you have left? The limiting reagent is the H2 in this case. So we can determine the amount of H2O produced: 1.5mol H2 * (2mol H2O/2mol H2) * (18.02g/mol H2O/1mol H2O)=27.03g of H2O The theoretical yield of H2O is 27.03g. To determine how much of the O2 we can relate it to the mols of H2O produced which is 1.5mols: 1.5mol H2O * (1mol O2/2mol H2O) * (31.99g/mol O2/1mol of O2)=23.99g of O2 are used. To figure out how much is left we have to determine the initial amount of O2 that we had: 5mol O2 * (31.99g/mol O2/1mol O2)= 159.95g O2 So 159.95g-23.99g=135.96g of O2 remains.

  2. .Sam.
    • one year ago
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    Fantastic

  3. Photon336
    • one year ago
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    Nicely done! Yeah a lot of people have questions about this.

  4. vera_ewing
    • one year ago
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    Awesome! Thanks! :)

  5. anonymous
    • one year ago
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    keep up the good work :)

  6. rvc
    • one year ago
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    :) we need more tutorials like this :) Thanks for your time and effort :)

  7. Zale101
    • one year ago
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    Keep it up!!

  8. taramgrant0543664
    • one year ago
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    Thanks everyone means a lot that you took the time to look at it!!

  9. Photon336
    • one year ago
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    Great job with this again

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