Stoichiometry Question?

Question

N2 + 3H2 --> 2NH3

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

How much of the excess reagent in Problem 1 is left over?

Please show your work.

Answer 1

you need to find which is limiting reactant
28gN2* 1mole N2/28.02g(2mole NH3/1mole N2)*(17.04gNH3/1moleNH3) =34.1gNH3
25gH2*(1 mole H2/2.02gH2) *(2 mole NH3 /3 mole H2)*(17.04 g NH3 /1 mole NH3) = 140.6gNH3
therefore N2 is limiting reactant cuz i have number g of NH3 is smaller than H2.
therefore gram of NH3= 34.1g
part b, following in part a
excess meant H2
34.1gNH3*(1moleNH3/ 17.04 g NH3) * (3 mole H2 /2 moleNH3) * (2.02 g H2/1 mole H2) = 6.06gH2
that is amount you used in this rxn. the leftover is 25-6.06=18.94gH2

good luck bro.

Answer 2

This is a limiting reactant problem.

You've already got a balanced equation, so the next step is to figure out how many grams NH3 would be produced if your 28g of N2 reacted completely.

Next, figure out how many grams NH3 would be produced if your 25g H2 reacted completely.

Compare the two amounts, as they are likely different. When this reaction actually occurs, you're going to create only as much NH3 as the smaller of the two numbers.

Do you see why? Since you have different amounts of reactant at the beginning, and they always react with each other in a certain ratio, you'll have excess of one of the reactants left over.

So, the reactant (N2 or H2) which created the smallest amount of product (NH3) is your limiting reactant--simply meaning it ran out before the other reactant had a chance to react completely. You call that one the "excess" reactant or reagent.

The next step should be pretty straightforward. Figure out how much of the excess reagent is left over. That is, when your limiting reactant reacted completely, how much of the other stuff is left? You should report your answer in grams.

Answer 3

N2 mass = 28
28 g N2 = 1 mol N2
1 mol N2 reacts with 3 mol H2 by balanced equation
3 mol H2 = 3 x 2 g/mol = 6 g H2 required
28 g N2 + 6 g H2 = 34 g NH3 produced
25 g H2 - 6 g H2 used = 19 g H2 left over