Friday
April 18, 2014

Homework Help: Honors Chemistry

Posted by Emily on Sunday, November 8, 2009 at 8:39pm.

"How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2?

N2 + 3H2 ---> 2NH3

==> 28 g N2 x (1 mol N2/14.0 g N2) x (2 mol NH3/1 mol N2) x ( 17.0 g NH3/ 1 mol NH3) = 68 g NH3

25 g H2 x ( 1mol H2/2.0 g H2) x (2 mol NH3/3 mol H2) x (17 g NH3/2 mol NH3) = 141.7 g NH3

So, 68 g NH3 can be produced because N2 is the limiting reactant and you will run out of it first.

To find how much excess reagent is left, we need to do the reaction backwards using the previous answer.

68 g NH3 x ( 1mol NH3/17.0 g NH3) x ( 3 mol H2/2 mol NH3) x ( 2.o g H2/1 mol H2) = 12 g H2

25-12=13 g H2 "

Is this explanation correct with how to solve this type of problem?? I asked this question a few days ago on this website but that was a different answer and explanation than what I got from another website.

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