Aluminum hydroxide reacts with sulfuric acid as follows:
2 Al1OH2...

Question

Aluminum hydroxide reacts with sulfuric acid as follows:
2 Al1OH231s2 + 3 H2SO41aq2¡Al21SO4231aq2 + 6 H2O1l2
Which is the limiting reactant when 0.500 mol Al1OH23 and
0.500 mol H2SO4 are allowed to react?

Accepted Answer

hey everyone in this example we have hydroxide of alkaline, earth metals that are basic in nature and react with strong acids like hydrogen chloride to produce alkaline, earth metal salts and water were given the following reaction. And we're told that when we have 4.50 g of barium hydroxide or barium to hydroxide Reacting with 5.6 g of hydrochloric acid, which one of the reactant is consumed before the other and identify which one it is. And so what this means by reactant that is consumed before the other. This is referring to the term which we should recall as the limiting reactant or the limiting re agent. So we want to verify that this given equation is balanced. And looking at the coefficients given in the equation, we can confirm that this is a balanced equation which is important to verify and we can pick any one of our products to see how much of that product is produced. So let's go ahead and see how much of our barium two chloride is produced from either one of our reactant. So we want to find the theoretical mass of barium two chloride From our first reactant. Barium two hydroxide, as well as the theoretical mass Of Barium two chloride produced from our second re agent, which is our hydrochloric acid. And because we have verified that we have a balanced reaction. Our next step is to also look at the ratio between barium two chloride and are very um to hydroxide as well as looking at the molar ratio between barium to chloride and hydrochloric acid. And this comes from our coefficients in our balanced equation. So we would see that we have a coefficient of one in front of our product Barium two chloride And then in front of our very um to hydroxide, we also have a coefficient of one. So we have a 1-1 molar ratio which we're just making note of for now and then for our barium chloride, we still have that ratio of one. And then in front of our hydrochloric acid. In our reaction we have a coefficient of two. So we can go ahead and fill in a ratio as a 1 to 2 ratio between barium to chloride and hydrochloric acid. So now that we have these molar ratios noted down, we can go ahead and get into the calculation. So starting off with the theoretical mass of our barium chloride produced from barium to hydroxide were given 4.50 g of barium to hydroxide from the prompt. And we want to go ahead and move from grams of Miriam to hydroxide into moles of barium to hydroxide. And so recalling from our periodic tables the molar mass of barium to hydroxide, we would see that we have for every atom and barium to hydroxide a mass of 1 71.34 g of barium hydroxide for one mole of barium to hydroxide. And now we're going to go into from moles of barium to hydroxide to get two moles of our product which is our barium, two chloride. And so this is the part where we plug in that molar ratio which we said we have one mole of barium to chloride produced from one mole of barium to hydroxide. Our last step in this conversion or calculation is to go from moles of barium to chloride to find how much grams we produce of barium to chloride. And this is where we want to plug in our molar mass, which from our periodic tables, we would see that for one mole of barium chloride, we have a molar mass equal to for every atom in barrie into chloride, 208.23 g. And so now we can go ahead and cancel out moles of barium chloride as well as moles of barium to hydroxide and grams of barium to hydroxide. Leaving us with grams of barium to chloride. And this is going to give us a value equal to 5. g of barium to chloride produced. This is our first theoretical produced mass from our first re agent barium to hydroxide. So now we want to go ahead and find the theoretical mass of our barium to chloride produced from our second re Agent hydrochloric acid and were given a mass of 5.65 g of our hydrochloric acid. So we're going to go ahead and go from grams of hydrochloric acid, two moles of hydrochloric acid. So we're calling from the molar mass of our, from our periodic tables of hydrochloric acid we would get for every atom in this compound and molar mass of 36.46 g for one mole of hcl. And so now we're going to move from one mole or sorry, from moles of hcl, two Mnolds of our product that we're comparing barium to chloride. And then lastly, we want to get from moles of barium to chloride, two g of barium two chloride. And so, plugging in that ratio or molar ratio between hydrochloric acid and buried into chloride. We stated from our balanced equation that we have one mole of barium chloride formed from two moles of hydrochloric acid And then plugging in that same molar mass of barium to chloride. We have 208.23g for one mole of barium to chloride. And we can go ahead and cancel out our units, leaving us with grams of barium chloride as our final unit, which is what we want And what we're going to get for a theoretical yield of barium chloride from our hydrochloric acid reagent is a value of 16.1 g of barium to chloride produced. And so what we can say is that 5. g of B A c l two is less than 16. g of be a cl two. And so therefore are limiting re agent, which is what we need to give the answer for For this question is going to be the reactant or the re agent that produced that lower mass of product. So the 5.47 g of grams. Sorry, the 5.47 g of barium to chloride was produced from our re agent vary um to hydroxide. And so we would say that therefore barry in two or that would be be a O H two is our limiting re agent. And this will be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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