Aluminum hydroxide reacts with sulfuric acid as follows:
2 Al1OH...

Question

Aluminum hydroxide reacts with sulfuric acid as follows: 
2 Al1OH231s2 + 3 H2SO41aq2¡Al21SO4231aq2 + 6 H2O1l2 
How many moles of 
Al21SO423 can form under these conditions?

Accepted Answer

Hi everyone for this problem. It reads consider the reaction between potassium hydroxide and sulfur dioxide below. Starting with a 1.55 moles of potassium hydroxide and 2.60 moles of sulfur dioxide, calculate the amount in moles of potassium cell fight produced. Okay, so we want to know the amount and moles of potassium sulfide produced from these two react ints. Okay. And so what we want to do here is determine which one is the limiting reactant. The limiting reactant is going to be the one that is consumed first. And that moller value that we calculate determines the theoretical yield and the problem. Alright. So we want to know from the two reactant, given how much potassium sulfide do they produce individually. So we're going to do to calculations here and then the one that is the lesser value is going to be our limiting reactant and that is going to be the theoretical yield of potassium sulfide produced. So let's go ahead and get started with our first reactant. Our first reactant is potassium hydroxide. So we want to go from we want to calculate how The amount of product produced from our potassium hydroxide and we know we have 1.55 moles of potassium hydroxide. Alright, so our goal here is to go from moles of potassium hydroxide to moles of our product potassium sulfate. Alright, so let's do that now. So we're going to set up a dimensional analysis here and we want to go from moles of reactant, two moles of product and we can do that using a mult Immel ratio. So looking at our reaction for every two moles of potassium hydroxide consumed, We have one mole of potassium sulfide produced. So that's our multiple ratio. So let's go ahead and write that here. So for every two moles of potassium hydroxide consumed, there is one mole of potassium cell fight Produced. Alright, so making sure units cancel properly are moles of potassium hydroxide cancel. And we're left with 0.775 moles of potassium soul fight produced from that amount of potassium hide dioxide. Alright, so we need to do the same thing. But with our other reactant now are sulfur dioxide. So we're told we have 2.60 moles of sulfur dioxide. So, our goal here is to go from moles of sulfur dioxide, two moles of potassium sulfide. We want to know how much product this reactant is going to produce. Alright, so let's go ahead and do the same thing. Look at our reaction. Okay, so looking at our reaction for every one mole of sulfur dioxide consumed, there is one mole of potassium cell fight produced. So let's go ahead and write that here. So, for every one mole of sulfur dioxide consumed, one mole potassium sulfite is produced. All right, So, our units of moles of sulfur dioxide cancel. And we're left with moles of potassium sulfite. This calculation yields 2.60 moles of potassium sulfate. So now we see what is the amount of product produced from each of these reactant. And like we said in the beginning, the reactant that produces the lesser amount is going to be our limiting reactant. So this is going to get used up first and our reactant, that is the larger number is going to be our excess reactant. Okay, so our limiting reactant is actually the theoretical yield in this problem. So this is going to be the answer. This is the amount of potassium sulfide produced. Okay, and so this is going to be our final answer because this is the theoretical yield. Once this, once this, once the moles of potassium hydroxide run out, that's it. That's why it's our limiting reactant, because there's going to be no more left. And so that makes this our theoretical yield and the final answer for this problem, I hope this was helpful.

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