"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.
You are correct that N2 is the limiting reagent but your numbers are not correct.
N2 + 3H2 ==> 2NH3
moles N2 = 28/28 = 1.0
moles NH3 = moles N2 x (2 moles NH3/1 mol N2) = 1.0 x 2 = 2.0
grams NH3 = 2.0 x 17 = 34 g.
moles H2 = 25/2 = 12.5 moles.
moles NH3 produced = 12.5 x (2 moles NH3/3 moles H2) = 8.33 moles NH3.
Therefore, N2 is the limiting reagent and 34. g NH3 will be formed.
Check my work.
For the reverse, you need not start with NH3; you may simply use the moles N2 you had.
1.0 mole N2 will require
1.0 moles N2 x (3 moles H2/1 mole N2) = 3.0 moles H2
grams H2 = 3.0 x 2 =6.0 g
etc.
can i get a thumbs up
the response for the original question:
"==> 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"
is all correct except that the atomic mass of nitrogen gas (N2) is 28.014g as nitrogen is a diatomic element, meaning you must multiply the atomic mass of N on the periodic table by 2. (14.007*2)
When worked out correctly, N2 is still the limiting reactant, but will produce around 34g of 2NH3.
"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.