Balancing Chemical Equations (Simplified) - Video Tutorials & Practice Problems

Balancing Chemical Equations (Simplified) Concept 1

in this series of videos, we're gonna take a look at balancing chemical equations when balancing. Always make sure the type and number off Adams on both sides of the arrow are equal. We're going to say in a balanced equation. The numbers that are in red are referred to as the coefficients. So we have to one and two as the coefficients for this balanced equation, and we're talking about type and number of atoms. We'd say that these coefficients get distributed so this, too, would get distributed to this hydrogen. There's already two of them, so two times to give me four hydrogen. This is just the one, so one distributed is one times two, which is two oxygen's on the other side. The two gets distributed, so it be two times the two hydrogen would give me four hydrogen. But there's also one oxygen here, so two times one gives me to oxygen's, so the types of atoms are the same on both sides. I have hydrogen and hydrogen oxygen and oxygen, and then the numbers off each are equal on both sides. Four. Hydrogen, four hydrogen to oxygen's to oxygen's. This is how we ensure a chemical equation is balanced. The types and numbers of each atoms are the same on both sides of the arrow. Now click on the next video and let's take a look at an example question where we go more refined at our approach to balancing a chemical equation.

Balancing Chemical Equations (Simplified) Example 1

here. We're told to write the balance equation for the following by inserting the correct coefficients in the blanks. All right. So, we have here, C four H 10 gas reacting with 02 gas to produce and give us a actual liquid and C 02 gas. All right. So step one, we have to set up a list for the elements that are reactant and another list for the elements that are products. All right. So on the reacted side we have four carbons, four carbons, 10 hydrogen, and two oxygen's on the product side, we have one carbon, two hydrogen, and then we have one oxygen here, but another two over here. So that's three oxygen's We can see the lists don't match. And that's because the equation is imbalanced. So step two, we're going to start from the top and go going down both lists determine how many of each element is present. Now here, this is important if a poly atomic iron is present on both sides, then treat as a single unit. So for example, if one side had one phosphate ion and the other side had three phosphate ions, you make sure that the this side here also has three phosphate ions in this particular equation. We don't have to worry about polly atomic ions, there are none. And then here step three, starting from the top and going down both of us begin balancing each element to ensure they match. Sometimes you may have a decimal fraction as a coefficient and so multiply the equation by two. All right, so we're going to do here is we're gonna see how many we have a beach, which we did. All right, so let's start doing this. We have four carbons here, but only one here. So I'm gonna put a four right here. That four gets distributed to the carbon, but it also gets distributed to the oxygen. So it's gonna be four times two, which is eight plus one more is nine oxygen's. Now now we keep going down the list, the carbons are balanced because they're both the same. You're going down the list hydrogen. Here, we have 10. And here we have to We need this site to also have 10. So, I'm gonna put a coefficient of five here. The five gets distributed to d hydrogen. So five times two gives me 10 hydrogen. But then it also gets distributed to the oxygen. So five times one is five oxygen's here. And remember, we still have another eight over here. So that's five oxygen's plus eight. Oxygen gives us 13 total oxygen's on the product side. Finally, We have to look at the oxygen's here, we have two. And here we have 13. So you have to think of a number that I can multiply by two, Which will give me 13. If I put a six here, that's six times two, which is 12. It's not enough. If I put a seven here, that's seven times two, which is 14. That's too much. So we've got to find something in between six and 7 And that would be 65 6, 5 times two gives me 13. Both of us MAC. Because now my equation is balanced. But remember step three says that we cannot have fractions or decimals as coefficients. When that happens, I multiply the entire equation by two. Okay, so this whole thing will get multiplied by two. So this coefficient of one gets multiplied by two. This coefficient of 6.5 gets multiplied by two. This coefficient of five gets multiplied right to And then this coefficient of four gets multiplied by two. So these are the new coefficients that we're going to have for each of these compounds. We're gonna have to c four H 10 gas plus 13 +02 gas gives me A produces H 20 liquid plus eight co 2 gas. So here this would represent the coefficients for each of the compounds and all we have to do is insert the correct coefficients in the blanks provided. Okay, so those would be the values that you put 2, 13, 10 and eight.

Write the balanced equation for the following by inserting the correct coefficients in the blanks.

• A. 1, 1, 3, 1
• B. 2, 3, 3, 1
• C. 1, 3, 6, 2
• D. 2, 3, 6, 1