Stoichiometry question?

Question

I am currently doing limiting reagent and excess part of it, and i have trouble, with calculating here is the question

N2 + 3H2 ---> 2NH3
How many grams of NH3 can be produced from the reaction of 28g N2 and 25g of H2?

I did 25g divide 2g = 12.5mols then divide by 3= 4.17mols
28g divide by 28g = 1 mol then divde by 1= 1 mol
so 1 mol is limiting and the 4.17 is excess
i use 1 mol times 2NH3 divide by 1 N2 = 2 mols x 17g = 34g of NH3
then the second part asks How much of the excess reagent in problem 1 is left over?
im assuming thats the 4.17mols, and i cannot find how to figure it out, becuase my answers does not match the answer on back of book..so please someone help, thanks

Answer 1

the answer is 19g H2.

Although your answer in the first question is right, your solution is kinda messy. For instance, the unit of your molecular weight is only in 'g'. It should be in 'g/mol'. This is important so that you won't get lost.

The excess reagent in problem 1, as you have solved, is H2 - which is equivalent to 12.5 moles. Since your limiting reagent, N2, only has 1 mole, this will only react to 3 moles of your H2 (see your balanced equation). So, 12.5 moles minus 3 moles is 9.5 moles H2 left. Multiply this with the molecular weight of H2, which is 2g/mol, you will get 19 grams of H2 left.

Answer 2

You are on the right track, but I think you need to be better organized.
You start with 1 mole of Nitrogen, which according to the reaction, will react with 3 moles of hydrogen. Your 3 moles of hydrogen is 6 grams of hydrogen. So, how much is left over? 19 grams. You really don't need to worry about how much NH3 you made; in fact, it seems to be tripping you up.