Balance the following equations: (b) Au2S31s2 + H21g2¡Au1s2 + H2S1g2

Video transcript

Hey everyone. Let's go ahead and balance this chemical reaction first. We're going to need to count the number of atoms we have on both sides. Starting off with our react inside, we have two of iron, three of sulfur and two of hydrogen in our product side. We have one of iron, one of sulfur and two of hydrogen 1st. Let's go ahead and balance out our iron. Now, if we add a coefficient of two prior to our iron in our product side, we end up balancing our irons perfectly. Now let's go ahead and look at our sulfur and our hydrogen. Now in order to balance out our cell furs, we're going to have to add a coefficient of three prior to our hydrogen sulfide. And when we do that, we change the number of sulfur and hydrogen we have in our product side. So we have three of sulfur and six of hydrogen since we have six of hydrogen. Now let's go ahead and look at our react inside and we can fix this by adding a three prior to our hydrogen gas, Which gives us six of hydrogen. So our final balanced chemical reaction is going to be like this. So I hope this made sense. And let us know if you have any questions