## Photon336 one year ago Basic Tutorial on balancing chemical equations

1. Photon336

To do any problem that involves stoichiometry you must first: Balance your chemical equation, and take the following into consideration. 1.You must make sure that the number of atoms is the same for each reactant and product. 2. And that your equation obeys the law of conservation of matter, because matter can neither be created nor destroyed and thus the mass of the products and reactants will be equal. 3. Same number of atoms on both sides of your equation.

2. Photon336

$H _{2} + N _{2} = NH _{3}$ Say we have the equation above, which represents the Haber process, Is this equation balanced? To determine this, we must look at the number of atoms of each reactant and product on both sides of the equation. IMPORTANT: you can't do any calculations without balancing your equation otherwise you'll be wrong every time.

3. Photon336

If you're doing this for the first time, ensure that you make a chart listing all the quantities that you have, in our case we have Reactants Products H = 2 N = 1 N = 2 H = 3 You can see why this isn't balanced right? We must balance them by adding coefficients in front of both the reactants and products: This is where our chart comes in handy: For nitrogen you have N2 on one side N on the product. 2 atoms on one side, one on the other. if you multiply the nitrogen on the product side by 2 you get N2; however this will also give you 6 Hydrogens: Because 2NH3 = 2N and 6H Like any equation what we do to one side effects what we do to the other so we must look at how many atoms we have on the reactant side. H2 + N2 = 2NH3 now nitrogen is balanced. why? Let’s then look at hydrogen it’s clearly not balanced. but we had 6 hydrogens on the product side WHILE ONLY 2 ON THE REACTANT SIDE. what must we do? we multiply the number of hydrogens by 3 because 3x2 = 6. Now 3H2 + N2 = 2NH3

4. Photon336

$N _{2} + 3H _{2} -> 2NH _{3}$ Now the result of all this is the following. Can you see why this is balanced? if so NOW we can WORK on any problem. sometimes you'll be asked to calculate how many grams of a product or something is produced, and they won't always balance the equation, so this step is fundamental and essential to getting these types of problems correct.

5. Photon336

Another important thing you must know in order to solve these kinds of problems: My peers and I have answered a lot of these questions: You MUST always calculate the molar mass of each compound in the reaction; and secondly, you MUST always take into account the molar ratio, and there's something important you must do in-order to get this and that's the following dimensional analysis. I will show you how to do both of these to the best of my ability. But first let's talk about molar ratios: Now the first step was essential, balancing the equation, because, we need to know the molar ratios of each of the quantities before we can do any calculations. N2 + 3H2 --> 2N3 $N _{2}+3H _{2 } -> 2NH _{3} Ratio of Nitrogen \to hydrogen N _{2}/3H _{2}$ Let's look at the ratio of nitrogen to hydrogen. we first look at the equation 3 moles of Hydrogen gas reacts with 1 mol of Nitrogen gas. If we take the molar ratio of Nitrogen gas to hydrogen gas that's a ratio of 1/3.

6. Photon336

Let's look at a problem: To completely convert 9.0 moles of Hydrogen Gas to ammonia gas how many moles of Nitrogen gas are required. The first thing we must do is to see what they are asking for and that is the moles of nitrogen gas that are required to make this reaction go. Now we have 9.0 moles of hydrogen gas, the question is how do we find what we are looking for, and that is the moles of nitrogen gas that we need to react. Well the hint is in the equation, we multiply the number of moles by the molar ratio.

7. Photon336

$9.0 moles H _{2} x (\frac{ 1mol N2 }{ 3 moles H2 }) = 3.0 Moles of N2$

8. Photon336

I want anyone reading this to pay close attention to how I set this problem up. whenever you are asked to find how many moles of something that's produced, you must always multiply by the molar ratio of those two quantities. now here's what i think is important. The quantity that you are looking for MUST ALWAYS BE IN THE NUMERATOR, while the quantity that you don't want in the denominator. In the problem above, we had 9.0 moles of Hydrogen gas multiplied by 1/3moles of hydrogen gas, and if you noticed the ratio was set up in a way that we had hydrogen in the denominator so when we multiplied the two quantities hydrogen canceled out, leaving us with 3 moles of nitrogen gas in the numerator.

9. Photon336

Let's do one more: let's say if we had 12 moles of nitrogen gas, how many moles of hydrogen gas would we need? Look at the molar ratio of N2 to H2 that's still 1/3. but however, we must find how many moles of H2 that we need to react. $12.0 mol N _{2} x (3H _{2}/moles N _{2}) = 36 moles of H _{2}$ You see how I set that up? we took the molar ratio, but we flipped it this time ensuring that nitrogen was in the denominator while hydrogen what we are looking for was in the numerator.

10. Photon336

Thanks for viewing this! hope this helped

11. anonymous

This is great. Very well organized and explains the process clearly!

12. Photon336

13. Photon336

This is kind of silly but this helps me to understand this stuff better. Let's look at it this way you have two things to consider. the first is the molar ratio and the second is the moles of a certain compound that you're given. the formula below will help you find the moles of any substance. what's in the brackets are the molar ratio Moles of compound you have x ( moles of compound you want /moles of compound you have) = moles of compound you want

14. sweetburger

I look at it in a similar way. I wish i had this way back 2 years ago when I was learning this. Very well done @Photon336. :)

15. Photon336

Thanks guys!

16. anonymous

@photon336 how do I search for specific tutorials when I need them?

17. Photon336

@cuanchi that's a good question! I saw people posting these so I decided to try one myself I'll find out.

18. nincompoop

balancing is the easiest portion of stoich

19. Photon336

If you've done it for a while then yes, it becomes intuitive.

20. sweetburger

Yes, but it is a huge backbone of many parts of math in chemistry @nincompoop.

21. rvc

good job! Very well done :) Excellent !!!!! Thanks :)

22. Photon336

@rvc your tutorial on kinetics inspired me! Haha you should do one on acids and bases next!

23. rvc

Aww Those words are beautiful :) You have put up in a nice manner :)