Consider this reaction 3H2(g) + N2(g) =2NH3(g) If 5.74g of H2 is reacted with 17.8 of N2. Find the limiting reactant and the theoretical mass of product of NH3

Question

I need step by steps clarification of getting these numbers. I can't seem to get the hang of this. explanation of where the numbers come from any why they are changing.

Answer 1

Stoichiometry can be tough. Until you realize it is kind of like making a recipe - let's say brownies. And you follow the directions on the box. Let's take a simplified brownie recipe from a box mix. Say it says you need 2 eggs (2E), 2 Tablespoons water (2T) and 1 package of mix (1M) to get 2 dozen brownies.

I could write the equation

2E + 2T + 1M = 2 dozen brownies.

Well what would happen if I only had 1 egg? I could halve the recipe. Then I would use 1 egg, 1 tablespoon water, 1/2 the package mix. I also could expect only 1 dozen brownies, not 2.

What would happen if I had 1 egg and a million packages of mix? Doesn't matter how much mix I have! Or water either - the egg is the limiting factor.

It gets a little complicated - but not really - when things are less obvious. Suppose I had 10 eggs, 10 tablespoons of water and 10 packages of mix. Now I have to figure out which one is the limiting factor. This can be done systematically until you get the hang of it.

Suppose I try to use all 10 of my eggs. According to my recipe I would also need 10 tablespoons of water and 5 packages of mix. I have enough of everything so this is a go! In fact, I will have mix leftover. According to the recipe, which calls for 2 eggs, since I am using 10 eggs, or 5 times the amount, - I should expect 5 times the product, or 10 dozen brownies. In a sense, the eggs, and the water, are LIMITING how many brownies I can make - because I run out of those things first.

If however, I decided to try and use all my mix - all 10 boxes of mix, - according to the recipe - I would need 20 eggs and 20 tablespoons of water. I don't have that! No go.

If you followed that - you can do this.

3H2(g) + N2(g) =2NH3(g)

This is my "recipe". Remember the coefficients, the numbers in the balanced chemical reaction, refer to the NUMBER OF MOLES.

You are given:

5.74g of H2 and 17.8 of N2

First off - grams is a useless thing to have. In chem, you always need moles. Convert to moles by using the formula (or molecular) mass.

H₂: (5.74g)(1 mole/2.02g) = 2.85 moles

N₂: (17.8g)(1 mole/28.0g) = 0.64 moles

That's what I am starting with.

According to the recipe/balanced chemical equation -

3 H₂ combines with 1 N₂

So if I have 2.85 moles of H₂, I would need (1/3)(2.85) = 0.95 moles N₂

THINK - I need to get a SMALLER number! So I DIVIDE by 3,

OR set up a fraction using the coefficients from the balanced chem rxn:

(2.85 H₂)(1 N₂/3 H₂)

Do I have 0.95 moles of N₂? No. I don't have enough N₂ to use up all my H₂. I ONLY have 0.64 moles of N₂ and I just calculated I need 0.95 moles of N₂ to use up all my H_2. So I CAN'T use up all my H₂.

Let's try that the other way:

If I use up all my N_2, all 0.64 moles, according to my recipe I need 3 times the amount, or 3x0.64=1.92 moles of H₂. I have plenty! I will RUN OUT of N₂. I can use it all! That is the LIMITING REACTANT - because it LIMITS the reaction by running out! The reaction stops when I run out of one of the "ingredients".

So I will use all 0.64 moles of N_2, and 1.92 moles of H₂ .

According to the recipe - i get twice as much NH₃ as N₂, so I should get 2 x 0.64 moles NH₃ = 1.28 moles NH₃.

The question is asking for MASS of the product, not moles, so now I have to convert that to grams.

Bake away!