Consider the reaction 2Al(OH)₃ + 3H₂SO

Question

Consider the reaction 2Al(OH)₃ + 3H₂SO₄ → x + 6y. what are x and y?

Accepted Answer

Welcome back everyone. We need to provide the balanced complete equation for the acid base reaction between peru vic acid and strontium to hydroxide. Let's begin by focusing on peru vic acid first H C three H 303. And were given the acquis label, recall that it's not on our memorized list of strong acids and is therefore a weak acid. And recall that based on its label as a weak acid, it's going to partially dissociate into the following ion products where we would have an equilibrium arrow. And on our product side we form one mole of our proton h plus as well as one mole of our Piru an ion C three H +3031 minus. And because we have an available proton on our product side, that means that peru vic acid has the ability of being a proton donor and is therefore are acid. In this case of the acid based reaction. As the prompt mentions now focusing on our second product being our strontium to hydroxide, we are also given an acquis label and recall that strong team to hydroxide is on our memorized list of strong basis, meaning that it will fully dissociate and we have a regular reaction arrow into the following ion products will reform one mole of our strontium two plus catalon and two moles of our hydroxide an ion. And because we have two moles of hydroxide products. We can see that due to the high electro negativity from the lone pairs and negative charge on hydroxide. It has the ability of being a proton except er and therefore our strontium hydroxide is definitely our base in the reaction. And specifically we are dealing with bronze, dead Lowry acids and bases. So now let's go ahead and write out how the reaction will form. Well we should recognize that our cat ian from our peru vic acid, which is our proton here is going to combine with the an ion, one mole of our an ion from our strong team to hydroxide. While our or rather an ion from terrific acid is going to combine with our Catalan from our second reactant circled here. And so we would form the following reaction where we have terrific acid Reacting with strontium two hydroxide in which we form the product strontium high piru Strong TM two pair of eight. Sorry. So that's why we have that subscript of two outside of the parentheses. And then our second product is going to be water. Now we need to make sure that this is a balanced equation. And so we need to compare our reactive side to our product side, in which we want to look at our atoms hydrogen, carbon, oxygen and strontium on both sides of the equation. So noting down what we have first on our reactant side we have three, 456 moles of hydrogen. And on our product side we have a count of three or rather 678 moles of hydrogen. Now for carbon we can count on a react inside a total of three moles of carbon. And on our product side we can count a total of six moles of carbon. Then on our reactant side for oxygen we can count a total of three, 4, 5 moles of oxygen. And then on our product side we can count a total of 6/7 moles of oxygen. Then moving on to strontium on our react inside, we can count one more. And on our product side we also have a count of one mole of strontium. So let's begin by balancing our non oxygen and non hydrogen atoms first. So let's try carbon. We see that we have only three moles on the reactant side. So to balance out with our product side where we have six moles we're going to place a coefficient of two in front of our peru vic acid, meaning that we would now have a total of six moles of carbon. Since we would multiply the coefficient of two by its subscript three and then that will change our number of hydrogen and oxygen on the reactant side. Where now we would have a total of two, two times three is six. So that's eight and then 9, 10. So now we have 10 moles of hydrogen on the reactant side. And then as far as oxygen on the reacting side, that would change to 6, 7, 8. And so we have now eight moles of oxygen on the reactant side. So so far we have balance strontium and carbon but we still need to balance out oxygen and hydrogen on both sides of our equation. And so recognize that if we place a coefficient of two in front of our two moles of water on the product side, Or in front of our one mole of product of water on the product side. That would actually change our number of moles of hydrogen on the product side to now four. And then plus six, which would give us 10 moles of hydrogen. And then that would change our oxygen to now two plus six, which would give us eight moles of hydrogen on the product side. And so now we can see that with the coefficients that we added. We now have a balanced chemical equation. And so for our final answer, we can highlight our entire balanced chemical reaction that is acid based between periodic acid and strontium to hydroxide. So what's highlighted in yellow is our final answer corresponding to choice be in the multiple choice. I hope this was helpful and let us know if you have any questions. See you in the next practice video

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