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Ch.4 - Chemical Quantities & Aqueous Reactions

Chapter 4, Problem 32

Sulfuric acid dissolves aluminum metal according to the reaction:

2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2( g)

Suppose you want to dissolve an aluminum block with a mass of 15.2 g. What minimum mass of H2SO4 (in g) do you need? What mass of H2 gas (in g) does the complete reaction of the aluminum block produce?

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Hey everyone, we're told that we want to dissolve an aluminum rod, weighing 1.5 kg. What is the minimum mass of sulfuric acid and grams required to react with aluminum? How much in grams of hydrogen gas is produced if the aluminum rod reacts completely And they provided us with our reaction. Now, before we answer this question, let's first see if our reaction is completely balanced out assessing this a bit further. We do need to balance this out and we can do so by adding a coefficient of two prior to our aluminum, a coefficient of three prior to our sulfuric acid and a coefficient of three prior to our hydrogen gas. Now our reaction is completely balanced out. So to answer this question, we need to use dimensional analysis. We want to take our 1.5 kg of aluminum and convert this into grams of sulfuric acid and also convert it into grams of hydrogen gas. So let's first determine the amount of grams of sulfuric acid. So starting off with 1.5 kg of aluminum, we're going to convert this into grams. Now we know that per one kg we have 10 to the third grams. Using aluminum's moller mass. We know that we have 26.98 g of aluminum per one mole of aluminum. And we can find this in our periodic table. Next, looking at the multiple ratios in our chemical reaction, we see that per two mole of aluminum, We have three mol of sulfuric acid. And this is coming from our coefficients right here. Next, using sulfuric acids molar mass, We know that we have 98.09 g of sulfuric acid Per one mole of sulfuric acid. Now, when we calculate this out and cancel out all of our units, We end up with a total of 8,180. g of sulfuric acid. Next, let's go ahead and convert our 1.5 kg of aluminum into grams of hydrogen gas. So taking the same steps, we're going to convert our kilograms into grams and we know that per one kg we have 10 to the third grams And we know that we have 26.98 g of aluminum per one mole of aluminum. Since this is aluminum's molar mass. Now looking at the multiple ratios between aluminum and hydrogen gas, we can see that we have two mole of aluminum, her three mole of hydrogen gas. And again, this is coming from our coefficients in our chemical reaction. And using hydrogen gasses, molar mass, we know that we have 2.2 g of hydrogen gas per one mole of hydrogen gas. Now, when we calculate this out and cancel out our units, We end up with a total of 168.46 g of hydrogen gas. And these will be our final answers. Now, I hope this made sense. And let us know if you have any questions
Related Practice
Textbook Question

Consider the balanced equation: SiO2(s) + 3 C(s)¡SiC(s) + 2 CO(g) Complete the table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product that forms. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that forms. Mol siO2 Mol C Mol SiC Mol CO _____ 1.55 _____ _____

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Textbook Question

Hydrobromic acid dissolves solid iron according to the reaction:

Fe(s) + 2 HBr(aq) → FeBr2(aq) + H2( g)

What mass of HBr (in g) do you need to dissolve a 3.2-g pure iron bar on a padlock? What mass of H2 would the complete reaction of the iron bar produce?

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Textbook Question

Hydrobromic acid dissolves solid iron according to the reaction:

Fe(s) + 2 HBr(aq) → FeBr2(aq) + H2( g)

What mass of HBr (in g) do you need to dissolve a 3.2-g pure iron bar on a padlock?

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Textbook Question

For each of the reactions, calculate the mass (in grams) of the product that forms when 15.39 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant. d. 2 Sr(s) + O2(g) → 2 SrO(s)

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For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutralize 4.85 g of the base. b. 2 HNO3(aq) + Ca(OH)2(aq)¡2 H2O(l ) + Ca(NO3)2(aq)
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Textbook Question

Find the limiting reactant for each initial amount of reactants. 2 Na(s) + Br2( g) → 2 NaBr(s) c. 2.5 mol Na, 1 mol Br2

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