Write the balanced chemical equation for each reaction. b. Solid iron(III) oxide reacts with hydrogen gas to form solid iron and liquid water.

Video transcript

All right. Hello everyone. So for this question, let's go ahead and write the balanced chemical equation or this reaction, solid iron three oxide reacts with hydrogen gas to form solid iron and liquid water. And here we have four different choices labeled A through D proposing different chemical equations for the reaction being described. So let's go ahead and get started. Now, the first thing I want to point out here is that the phrase to form in the description of this chemical reaction essentially represents the arrow separating the reactants from the products in the equation itself. So in terms of our reactants, we have iron three oxide, which if you recall has a chemical formula of fe 203, it is an oxide in which iron has a charge or an oxidation state of positive three and it exists as a solid. Now iron three oxide is going to react with hydrogen gas to create art products. Now, hydrogen being a diatomic exists as H two in its elemental or gaseous state. And now that we have the chemical formulas for our reactants, I'm going to go ahead and add the arrow, the reaction arrow to separate reactants from products. Now, the first product I see in the description provided is solid iron. So because it is a solid element, I would simply represent it as its chemical symbol or fe in our last product is liquid water. So H2O has a liquid. Now it's worth mentioning that I'm leaving some spaces before the chemical formulas of all of these compounds. And the reason for this is because our next step is to balance this chemical equation. In other words, making sure we have the same quantities of all atoms present on either side. So on the bottom of this chemical equation, that's currently unbalanced. I'm going to go ahead and keep track of all of my elements. In this case, I have iron, I have oxygen and I have hydrogen and I'm going to create a table for each of these elements, keeping score or keeping track of how many of each there are on the reactant side. And on the product side now starting off with iron, I can see that I have two atoms of iron on the reactant side in iron, three oxide and only one. So just to clarify, right, we have two atoms of iron on the left side, on the reactant side and only one on the product side, moving on to oxygen, we have three on the reactant side and only one on the product site. And for hydrogen, we have two on the reactant side and two on the product side. Now, it just so happens that hydrogen is the only atom that is currently balanced as the chemical equation is currently written because we have the same amount of hydrogen atoms on either side of the equation itself. So in this case, our task is to balance all of the elements present in this chemical reaction. So let's go ahead and start by balancing out oxygen first because there are three oxygens on the reactant side and only one on the product side. A coefficient must be added to the product side to increase the number on the right. So in this case, the product that contains oxygen is water. So I would place a coefficient of three in front of the chemical formula of water. And so by doing this right, I've increased the number of oxygen atoms on the product side to three. However, I've also affected the number of hydrogen atoms and to be more specific, I've increased the number of hydrogen atoms on the product side up to six. So by balancing oxygen, I've unbalanced hydrogen. But before we go ahead and address that, in other words, before we go ahead and rebalance hydrogen, let's go ahead and balance iron first or at least before that. Right. Again, we have less iron atoms on the product side, which means the product side is where a coefficient must be added. And because the ratio is 2 to 1, I'm going to place a coefficient of two in front of elemental iron on the right side. And this brings up the number of iron atoms on the product side, up to two, once more or rather up to two. So now iron is balanced, but let's not neglect hydrogen. Right. At this point, we only have two hydrogens on the reactant side and six on the product side. So this means a coefficient must be added in front of hydrogen gas on the reactant side. Now, in this case, on the left side, I must bring up the number of hydrogens 26. And in this case, I have two, right. So six divided by two equals three, which means that three is the coefficient that we must place in front of hydrogen gas. And so this brings up the number of hydrogen atoms up to a six on the reactant side. And now the equation is fully balanced. So here is our final answer. We have fe 203 reacting as a solid with three equivalents of hydrogen gas or H two to produce two equivalents of solid iron and three equivalents of liquid water. And this matches with option D in the multiple choice. So option D is our final answer and there you have it. So if you stuck around until the end of this video, thank you so very much for watching. And I hope you found this helpful