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Ch.3 - Mass Relationships in Chemical Reactions

Chapter 3, Problem 41

Balance the following equations. (a) VCl3 + Na + CO --> V(CO)6 + NaCl

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Welcome back everyone in this example, we need to write a balanced equation by providing the correct coefficients where we have the reaction of vanadium Penta oxide reacting with our second reactant vinyl chloride, where we produce our first product vanadium oxy tri chloride and then our second product sulfur dioxide. So let's write out this equation bigger. So it's clear V 205 plus our vinyl chloride, C O S O C L two yields vanadium oxy tri chloride plus sulfur dioxide. Recall that before balancing our equation, we need to compare our reaction side to our product side. So let's draw this line here to separate both sides. We have vanadium, oxygen, sulfur and chlorine on both sides of our equation to worry about. And on our reactant side we can count a total of two of our vanadium adam vanadium atoms. We then can count a total of 56 of our oxygen atoms. We then can count for chlorine, a total of two atoms on the reactant side. And then for sulfur we can count just one of our sulfur atoms on the reactant side on our product side we count also one vanadium atom. For oxygen, we count a total of 23 oxygen atoms. For chlorine, we count a total of three chlorine atoms And for sulfur we count a total of one sulfur atom on the product side. So we're going to focus on balancing vanadium first by placing a coefficient of two in front of our vanadium oxy tri chloride on the product side, which is now going to give us to vanadium atoms. We're now going to have two oxygen atoms plus two, which would give us four oxygen atoms now on the product side. And then we're also going to change our chlorine atoms on the product side two now three times two, which is going to give us six chlorine atoms on the product side. So now balancing out our chlorine atom, we want to go ahead and place a coefficient of three on our product side, in front of our Fiona alkaloid reactant. So now for our final chloride reactant, we have now changed our sulfur atoms on the reactant side, two now three, Our oxygen atoms on the reactant side too, now eight and our chlorine atoms on the reacting side to now six. And now that we've balanced chlorine, let's focus on balancing our sulfur atoms. So to balance our sulfur atoms, we're going to place a coefficient of three in front of our vanadium Penta oxide reactant and sorry about that, not our vanity and Penta oxide at three in front of our sulfur dioxide because we want to bounce out sulfur as we stated. So this would change our sulfur on the product side, two now, three moles of sulfur, which is also going to change our oxygen to now six moles of oxygen will rather seven since we have we have three times are two oxygen atoms at six. And then we have two added to that, which would give us eight actually, So we have eight oxygen atoms now on our product side. And now with this change, we can confirm that we now have a balanced reaction. And so for our final answer, our balanced equation is going to be this equation with our added coefficients. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.