What is the minimum amount of 6.0 M H₂SO₄ necessary to produce 25.0 g of H₂(g) according to the reaction between aluminum and sulfuric acid? 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

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All right. Hi, everyone. So this question is asking us what is the minimum amount of 6.0 molar H two. So four necessary to produce 25.0 g of hydrogen gas according to the reaction between aluminum and sulfuric acid. So that's two moles of aluminum react with three moles of sulfuric acid to produce one mole of aluminum sulfate and three moles of hydrogen gas. So here we also have four different answer choices labeled A through D proposing different volumes of the sulfuric acid solution. So let's go ahead and get started. Now, in this case, we're going to have to use dimensional analysis to relate the mass of hydrogen gas produced to the amount of sulfuric acid necessary when this reaction begins. Right. So in this case, we're going to use quite a few conversion factors. So before we discuss how that works, I want to just point out that the amount of hydrogen gas or H two is provided in units of grams. And in this case, we're provided with a balanced chemical equation that relates the moles of reactants to the moles of products. So our first conversion factor would have to be the molar mass to go from grams of H two moles of H two. And so recall that the molar mass of hydrogen gas is equal to 2.016 g per month. So let's go ahead and get started with our dimensional analysis. The first value that should always be written is the information that you are given, right? So first, I'm going to start with my 25.0 g of H two. And the first thing I'm going to do is convert from grams of H two to moles of H two using the molar mass. Now, in this case, I'm going to treat conversion factors as if they were fractions with a numerator and a denominator. The idea is to always make sure that your starting units cancel out and you're left with a new set of units. In the case of this particular first step, the idea is to cancel out units of grams and convert that into moles. This means that grams should cancel out when applying my molar mass. So in this case, when applying the molar mass of H two units of moles should go in the numerator and grams should go in the denominator because this ensures that my grams of H two cancel out to yield moles. So now that I have moles of hydrogen gas, I can use the mole ratios provided in the balanced chemical equation to convert from moles of H two to moles of sulfuric acid because recall that were trying to find how much sulfuric acid is necessary to produce 25.0 g of H two. Now, according to the balanced chemical equation, right, three moles of sulfuric acid correspond with three moles of H two. So those are the exact values that I'm going to use in my next conversion factor. Now, recall that right now our units are moles of H two. This means that in my second conversion factor that uses the mole ratios, moles of H two should go in the denominator because once again, my old units have canceled out, specifically moles of H two have canceled out. And now my answer is going to be referring to the moles of sulfuric acid. So at this point, we've calculated how many moles of sulfuric acid are necessary to produce 25.0 g of hydrogen gas. So now we're going to use the molarity provided or the molar concentration of sulfuric acid to find the volume necessary for the reaction. So my molar concentration here is going to be my third and final conversion factor. In which in this case, there are 6.0 moles of H two. So four per every 1 L of solution. So when he's using my third conversion factor, my moles of sulfuric acid should go in the denominator to make sure that those units cancel out, which means that leaders should go in the numerator moles of H two. So four cancel out. And now we can go ahead and calculate our final answer which will be expressed in units of liters of H two. So four. So after multiplying all numerators and dividing by the denominators and rounding to two significant figures, our answer ends up being 2.1 L of H two. So four and this corresponds to option D in the multiple choice and there you have it. So it, if you watch this video all the way up to the end. Thank you very much for watching. I appreciate it. And I hope you found this helpful.

What is the minimum amount of 6.0 M H₂SO₄ necessary to produce 25.0 g of H₂(g) according to the reaction between aluminum and sulfuric acid? 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

Verified Solution

Key Concepts

Stoichiometry

Molarity

Limiting Reactant

What is the minimum amount of 6.0 M H2SO4 necessary to produce 25.0 g of H2(g) according to the reaction between aluminum and sulfuric acid? 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

Verified Solution

Key Concepts

Stoichiometry

Molarity

Limiting Reactant

What is the minimum amount of 6.0 M H₂SO₄ necessary to produce 25.0 g of H₂(g) according to the reaction between aluminum and sulfuric acid? 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)