Balance the following equations: (d) KOH1aq2 + H3PO41aq2 ¡K3PO41aq2 + H2O1l2

Video transcript

Hey everyone. We're asked to balance the given chemical reaction first. Let's go ahead and count the number of atoms we have on both sides. Starting off with our react inside. We have one of nickel, one of hydrogen, one of nitrogen And three of oxygen. Looking at our product side, we have one of nickel as well, two of hydrogen, three of nitrogen and nine of oxygen only. Our nickel is balanced out. Let's first start off by balancing out our hydrogen. So if we add a coefficient of two prior to our nitric acid, we then are able to change our hydrogen into two, our nitrogen into two as well, And our oxygen's into six. Since we still need to change our nitrogen and oxygen's, this means we have to change The coefficient that we just inputted, which is R two. So instead of two, we can then go ahead and input a four prior to our nitric acid. And when we do that, we change our hydrogen into four, Our nitrogen into four as well. And our oxygen's into 12. Now that we've done this, we can adjust accordingly and look at our product side, we can add a coefficient of two prior to our water. And when we do this, we end up with four of hydrogen and our oxygen's changed from 9 to 10. Lastly, all we need to do is fix our nitrogen and our oxygen's and we can do that by fixing our nitrogen dioxide. So when we add a two prior to nitrogen dioxide, We end up with four of nitrogen And 12 of oxygen. So our balanced chemical reaction is going to be as such. So I hope this made sense and let us know if you have any questions.