The following equilibria were measured at 823 K:
CoO1s2 + H21g2 Δ...

Question

The following equilibria were measured at 823 K:
CoO1s2 + H21g2 ΔCo1s2 + H2O1g2 Kc = 67
H21g2 + CO21g2 ΔCO1g2 + H2O1g2 Kc = 0.14
(d) If the reaction
vessel from part (c) is heated to 823 K and allowed to
come to equilibrium, how much CoO1s2 remains?

Accepted Answer

Hello everyone today. We are being given the following problem. And to solve for it, It says that 700 Kelvin, the reaction in this reaction given to us has a K. c value of 52. A 15 g sample of iron was placed in a two liter reaction vessel along with .15 molar of water in the gaseous form and allowed to reach equilibrium at equally determined the mass of iron that remains. So the first thing we're gonna do in a question like this is right out our K. C expression, which will be K. C. Is equal to our concentration of our products over the concentration of our reactivates our products. In this case, we only want to pay attention to the gaseous forms, the gaseous molecules. And so that's just going to be our hydrogen gas or H2. And since we have a four as a coefficient that is going to become our exponent. And in the reactant we only have one molecule that's in the gaseous form and that's water. So we're going to write Htoo and brackets. And since that has a coefficient of four, we're going to make that exponent for as well. And so to solve for our moles of iron and moles of H20. That we need for our eventual ice table. That will need to construct will simply write out that the moles of iron is equal to the mass given to us are 15 g, we're going to multiply that by the molar mass. So we're going to say that one mole of iron is equal to the molar mass, which is 55.85 grams. And this is according to the periodic table When these units cancel out, we are left with 0.269 moles of iron. We're gonna do the same thing for our water. And so our water, we actually don't perform any calculations because it was given to us in the question stem. And so now we can move on to our ice table. So as usual, we're gonna write out our chemical equation. So we have our iron and the solid form reacting with our water and the gas form in an irreversible and reversible reaction to give us this ferrous oxide in the solid form as well as molecular hydrogen or just hydrogen gas. So we're gonna write R I C. E. The I stands for the initial concentration, the C. Is the change and the is the equilibrium or essentially the I minus R. C. So our initial Moles of our iron is going to be 0.269. And for water it's going to be 0.15. Of course we don't have an initial concentration of our products. So it's just gonna be zero. Now for our change, we know that this would usually be minus X since we're taking away from the reactant side. But since there's a three in front of that iron, it's going to be minus three X. And for water, since it's four as a coefficient we're gonna say minus four X. And then we just simply going to add X to both sides or just to the ferrous oxide. But to the oxygen to the hydrogen gas, we're going to add four X. Since there's a four coefficient in front of the hydrogen gas, E. Is going to be the equilibrium. So the I minus R. C. And that's going to the 2.69 subtracted by three X. For water, it's going to be 30.15 minus four X. And then ferris oxide is just going to be X and four. How you doing gas? It's just going to be X for X. We can then apply this to our K. C reaction. So we're going to say that our Casey has given to us in the question Stem was 52. And we're going to equal that with our variables that we just sold for. So as usual as before we're going to write our concentration of products over reacting. And so we're going to make note that this is going to be four X. And we're going to give that four X. That fourth power. Since there is a four coefficient, we aren't paying attention to solid. So that's why that ferrous oxide isn't not included four hour water here. That's going to look like 0.15 subtracted by four X. And also gets a four as an exponent. Since that is a coefficient simplifying this, we can first take the fourth root of both sides and that's going to give us 2.685 is equal to four X over 0.15 minus four X. With some algebra. We can solve for X to give us 0.034 moles. And so lastly we must find the number of moles remaining, so the number of moles of iron that our remaining. And so what we want to do is we want to take our 0.269 moles are starting material and we wanted to attract by how much we used up. And so note that we had a three in front of the original iron in our reaction to three times our Number of moles that we just solved for 0.034. And when we do that we get 0.167 moles. However, the question asked for this value in grams. So we must find the mass of the iron remaining from the moles. And to do that, we simply take the Molds that we have .167 moles of our iron. We're gonna multiply that by the molar mass of iron itself to get rid of these moles. And so that's going to be our 55.85 g of iron over one mole of iron And our units are going to cancel. Like usual, we're gonna be left with 9.3 grams of iron remaining. I hope this helped, and until next time.

The following equilibria were measured at 823 K: CoO1s2 + H21g2 ΔCo1s2 + H2O1g2 Kc = 67 H21g2 + CO21g2 ΔCO1g2 + H2O1g2 Kc = 0.14 (d) If the reaction vessel from part (c) is heated to 823 K and allowed to come to equilibrium, how much CoO1s2 remains?

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