When iron rusts, solid iron reacts with gaseous oxygen to form solid iron(III) oxide. Write the balanced chemical equation for this reaction.

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Hi everyone today. We have a question that tells us when solid copper reacts with oxygen gas. The product is solid copper, one oxide, and we need to write the balanced chemical equation. So we have copper solid reacting with oxygen, gas and oxygen is always in two. And then that's forming Copper one oxide. So we need to look at our charges and see what that's going to be. So we have copper with a plus one charge and oxygen with a -2 charge. And we need to crisscross those. So that would give us copper too oxygen. So that is our product. And now we need to balance this. So let's write out what we have. We have one copper on our reactant side And we have two oxygen And on our product side we have two copper and one oxygen. So our easiest step here is going to be to put a two in front of our Copper oxide and that gives us four copper and two oxygen. So our oxygen's are fine. Now we just need to fix our copper so we can put a four in front of our copper on the reactant side and that will give us four copper. And here is our balanced chemical equation. Thank you for watching. Bye

When iron rusts, solid iron reacts with gaseous oxygen to form solid iron(III) oxide. Write the balanced chemical equation for this reaction.

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Key Concepts

Chemical Reactions

Balancing Chemical Equations

Oxidation-Reduction Reactions