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anonymous

  • one year ago

If 62.7 g N2 react with 23.8 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over? Unbalanced equation: N2 + H2 → NH3 Be sure to show all of your work. @photon336

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  1. Photon336
    • one year ago
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    Do you know how to balance the equation?

  2. taramgrant0543664
    • one year ago
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    The first thing we have to do is balance this so we get N2+3H2---->2NH3 Then we can determine our limiting reagent: mol of N2= 62.7g * (1mol/ 28g/mol)=2.24mol mol of H2=23.8g * (1mol/2.02g/mol)=11.78 mol Now we need to take the ratio into consideration: 1 mol of N2 makes 2 mol of NH3 so the ratio is 1:2 This means that since I have 2.24 mol of N2 I can make 4.48 mol of NH3 3 mol of H2 makes 2 mol of NH3 so the ratio is 3:2 This means that since I have 11.78mol go H2 I can make 7.85 mol of NH3 So my limiting reagent is N2 as it makes lessNH3 so that means that H2 is the excess reagent. To determine how much is left you can do the following: 2.24 mol N2 * (3mol H2 /1 molN2) * (2.02g/mol H2 / 1mol H2)=13.57g H2 was used To find out how much was left you take 23.8-13.57= your answer

  3. Photon336
    • one year ago
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    Well explained Tara

  4. taramgrant0543664
    • one year ago
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    I made a tutorial a few days ago if you are still having trouble understanding it has two examples that I walked through: http://openstudy.com/study#/updates/559c49b4e4b0f93dd7c278ad

  5. anonymous
    • one year ago
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    @photon336 So the balance equation would be N2+3H2 -->2NH3?

  6. Photon336
    • one year ago
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    @lovinggod786 correct! So the next thing you need to do is to figure out how many moles of each that you have. Grams of N2 x (1mol N2/28g) = mol N2 Grams of H2 x (1 mol H2/2g) = mol H2 This will give us the number of moles that we HAVE not the number of moles we NEED to run the reaction.

  7. anonymous
    • one year ago
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    would it be 10.23 grams of h2 or n2 left over?

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