No, no, and no.
28 g N2 = ??moles = 28/14 = 2 moles N2.
2 moles N2 x (2 moles NH3/1 mole N2) = 2*2/1 = 4 moles NH3.
grams = moles NH3 x molar mass NH3 = about close to 70 g.
There is NO NH2 produced. It isn't even in the equation.
How much H2 remains un-reacted?
You have 25 g H2 which is 25/2 = 12.5 moles. How much H2 is used? That is
2 moles N2 x (3 moles H2/1 mole N2) = 2*3/1 = 6.0.
So you must have 12.5-6.0 moles left over which you can convert to grams if that is desired.
N2+ 3H2--> 2NH3
How many grams of NH3 can be produced from the reaction of 28 g N2, and 25 g of H2??
I got 34g NH3 and 140 NH2
2. How much of the excess reagent in problem 1 is left over?
1 answer