anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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JFraser
  • JFraser
How many moles of N2 and H2 do you have?
anonymous
  • anonymous
62.7 g N2 * mol/28g = 2.24 mol 23.8 g H2 * mol/2g= 11.9 mol ?????
JFraser
  • JFraser
Ok, now what molar ratio does the balanced reaction require? How many moles of H2 will react with 1mol of N2?\[N_{2} + H_{2} \rightarrow NH_{3}\]\[\frac {mol N_{2}}{mol H_{2}} = ?\]

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More answers

anonymous
  • anonymous
N2 and H2 they both have 1:1 molar ratio
JFraser
  • JFraser
The reaction's not balanced, the ratio has to be: \[N_{2} + 3H_{2} \rightarrow 2NH_{3}\] 1 mole of N2 requires 3 moles of H2 in order to fully react. You have 2.24mol of N2 available, so how much H2 is needed to fully react with that, if 1mol of N2 requires 3 mol of H2?
anonymous
  • anonymous
Yeah I balanced it to- N2 + 3H2 = 2NH3
JFraser
  • JFraser
since 1 mole of N2 needs 3 moles of H2, how much H2 do you need if you have 2.24moles of N2?
anonymous
  • anonymous
So wait, how do I find H2?
JFraser
  • JFraser
the molar ratio is 1N_2 : 3H_2. 2.24 moles of N2 will require three times that much H2 because the balanced reaction says so
anonymous
  • anonymous
So do I multiply 2.24 by three?
anonymous
  • anonymous
H2 has 11.9 mol???
JFraser
  • JFraser
you have 11.9mol of H2, but only need 3*2.24mol of H2, so the N2 will run out first. The N2 is the "limiting reactant". The limiting reactant dictates how much product you can make. The limiting reactant will be used up entirely, so how much NH3 will be made if you use all 2.24mol of N2?

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