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 
the excess reactant remain after the completion of the reaction?

Accepted Answer

hey everyone in this example, we're told that although zinc hydroxide is insoluble in water, it reacts in the below reaction. And we need to go ahead and calculate the number of moles of excess re agent when the reaction is complete. So our first step is to check out the coefficients in this given reaction and make sure that everything is balanced. And based on our coefficients, we can confirm that we do have a balanced reaction. Our first step is to find our moles of our first reactant, zinc hydroxide. And so in order to do so, we're going to take that mass that we said from the prompt as 3. g of zinc hydroxide. And we're going to convert to moles. So we want grams of zinc hydroxide to cancel out in our denominator so that we can get two moles of zinc hydroxide. And so we would get from our periodic tables, the molar mass of zinc hydroxide as .4 - four g of zinc hydroxide for one mole of zinc hydroxide. So this allows us to cancel our units of grams. Leaving us with moles of zinc hydroxide and this is going to give us a value equal to 0.0347 moles of zinc hydroxide. We want to follow the same steps to find our moles of hydrochloric acid, our second reactant. So according to the problem, we're reacting with 10.50 g of hcl. And so we want to cancel out grams of hcl to get two moles of hcl. So using our molar mass for hydrochloric acid from our periodic tables, we see that we have 36.46 g of hcl for one mole of hcl. We're able to now cancel out our units of grams, leaving us with moles of hcl and this gives us a value equal to 0.288 moles of hcl that we're reacting with. As you can see, we have this amount of moles of zinc hydroxide which is smaller than the moles of hydrochloric acid. And so this means that our moles of zinc hydroxide is going to be our limiting free agent because it has the smaller amount of moles as a reactant, meaning that our hcl is in excess. So we want to find out how many moles of hcl is consumed And to find that we're going to take our moles of our zinc hydroxide, which again we set is 0.0347 moles of zinc hydroxide. And we want to look at the ratio between our zinc hydroxide And our hydrochloric acid. And this is going to come from our balanced equation above. So we see we have a coefficient of one in front of our zinc hydroxide and a coefficient of two in front of our hcl and this means we have a 1-2 ratio. And so this allows us to go from moles of zinc hydroxide to the value, we want to calculate moles of hcl. So we can plug in that ratio where we said we have one mole of zinc hydroxide reacting with two moles of H C. O. And so this allows us to cancel out our units of moles of zinc hydroxide to give us our moles of hcl that is consumed in this reaction. And we're gonna get a value of 0.0694 moles of Hcl consumed. And so because we know that HDL is in excess because it's not the limiting reactant, our zinc hydroxide itis, we want to find the hcl, the moles of hcl excess that is remains or that is remaining. And so what we would do is take our moles of Hcl which we calculated above that should be reacting as .288 moles. So this is what reacts in our equation And we're going to subtract this from the amount of hcl that we found above. That is consumed by our limiting reagent zinc hydroxide. And so we said that that is 0. moles of hcl. And this difference gives us the excess hcl left over in our reaction as 0.219 moles of hcl left over. And this is going to be our final re agent or sorry, our final answer to this example as our excess free agent hcl left over when the reaction is complete. 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.

Aluminum hydroxide reacts with sulfuric acid as follows: 2 Al1OH231s2 + 3 H2SO41aq2¡Al21SO4231aq2 + 6 H2O1l2 How many moles of the excess reactant remain after the completion of the reaction?

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