The commercial production of nitric acid involves the following chemical reactions: 4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O(g) 2 NO(g) + O 2 (g) → 2 NO 2 (g) 3 NO 2 (g) + H 2 O(l) → 2 HNO 3 (aq) + NO(g) (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 M aqueous solution of nitric acid? Assume all the reactions give 100% yield.

Hi everyone for this problem, we're told the following series of reactions show the production of sulfuric acid from the extraction process of lead and were given the series of reactions below. If you want to obtain 200 liters of 2000.175 moller acquia solution of sulfuric acid. We're being asked to calculate the mass Of lead to sulfide in g. So we want the mass of lead to sulfide in g. If you are required to start with assuming that each of the steps has a 100% yield, including the second step. Okay, So what we're told is we have 200 liters of 2000.175 molar sulfuric acid and we want to go from sulfuric acid to mass of lead to sulfide. So because we're given the malaria t of sulfuric acid, male arat E. Is equal to moles over leader. Okay, so we know that we have 0.175 moles of sulfuric acid for every one leader of solution. Okay, so let's start off with what we're given and work towards what we want. So we know the volume of sulfuric acid. So let's start there. We have to We have 200 leaders of sulfuric acid. Okay, so we want to go from sulfuric acid. So volume of sulfuric acid, two moles of sulfuric acid. We can use that using the malaria T. They gave us in the problem. So we know that in The mill arat. E. is there 0.175 moles of sulfuric acid for every one liter of solution. Okay, so that means our volume of sulfuric acid cancels. So we want to go from sulfuric acid to lead to sulfide. So we need to look at the mole to mole ratio of everything in our problem and use that to go from what we're given to what we want. So let's take a look at some of our multiple ratios here. So we can see that we have, for the first reaction, there's two moles of lead to sulfide for every two moles of S. 02. For the second reaction, two moles of S. 02 for every two moles of S. 03. For the third reaction there's one mole of S. 03 for every one mole of H two S 207. And for the last reaction there's one mole of H two S 207 for every one mole of H two S. 04. Okay, so let's go ahead and use those multiple ratios. So for the first one we want to go from moles of sulfuric acid that H two S. 042 moles of H two S 207. Okay, so we see that we have a 1 - two Molar ratio. That's the last reaction. So for every one mole of H two S 207, we have two moles of H. Two S. 04. Okay, so now our moles of H two S. 04 cancel. And we're in moles of H two S 207. So we want to go from moles of H. Two S. 207 to most of S. 03. And our multiple ratio here is one and that is the third reaction. So we have for every one mole of S. 03, we have one more of H. Two S. 207. Okay so here our moles of H. Two, S. 207 cancel. And we're in moles of S. 03. So now we want to go from moles of S. 032 moles of 02 and that is a 222 Molar ratio. So that is our second reaction. So when we do that we have for every two moles of S. 02 There is two moles of s. 0. 3. Okay and now we can cancel out our moles of S. 03 and we're in moles of S. 02. So the last thing we want to do is go from moles of S. 022 moles of lead to sulfide, which is what we want. So the multiple ratio here is 2 to 2. So that is our first reaction. So for every two moles of lead to sulfide there is one more or two moles of S. 02. Okay, so now our moles of S. O. To cancel and we're in moles of lead to sulfide. Remember the question was asking us to calculate the mass of lead to sulfide. So we needed to go from volume of sulfuric acid, Two moles of lead to sulfide. So let's go ahead and calculate this. And when we do calculate it, we're going to get an answer of 17.5. So I'll draw this year, we have 17.5 moles Of lead to sulfide. Okay, so, but the question asks us for mass and mass needs to be in g. So the next thing that we're going to want to do is go from moles of lead to sulfide. So let's rewrite that here. So we have 17.5 moles of lead to sulfide. And we want to go to grams and we can do that using the molar mass of lead to sulfide using the periodic table. And so one mole of lead to sulfide using its molar mass. We find it weighs 239 .266g. Okay, so now our moles cancel, moles have led to sulfide cancel. And we're left with grams. This is the unit that we want because we want mass in grams. So once we go ahead and calculate that out, we get a final answer of 4187 g of lead to sulfide. And this is our final answer. This is the mass of lead sulfide in grams required to start with 200 liters of 2000.175 molar solution of sulfuric acid. That's the end of this problem. I hope this was helpful

Ch.4 - Reactions in Aqueous Solution

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Brown 14th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 96c

Chapter 4, Problem 96c

The commercial production of nitric acid involves the following chemical reactions:
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
2 NO(g) + O₂(g) → 2 NO₂(g)
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
(c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 M aqueous solution of nitric acid? Assume all the reactions give 100% yield.

Verified step by step guidance

Step 1: Calculate the moles of nitric acid (HNO₃) needed using the molarity formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Given that the volume is 1000.0 L and the molarity is 0.150 M, calculate the moles of HNO₃.

Step 2: Use the stoichiometry of the third reaction to find the moles of NO₂ required to produce the calculated moles of HNO₃. The balanced equation shows that 3 moles of NO₂ produce 2 moles of HNO₃.

Step 3: Use the stoichiometry of the second reaction to find the moles of NO required to produce the calculated moles of NO₂. The balanced equation shows that 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.

Step 4: Use the stoichiometry of the first reaction to find the moles of NH₃ required to produce the calculated moles of NO. The balanced equation shows that 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO.

Step 5: Convert the moles of NH₃ to grams using the molar mass of NH₃ (17.03 g/mol). Multiply the moles of NH₃ by its molar mass to find the mass in grams.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equations. It allows us to determine the relationships between the quantities of substances involved in a reaction. In this case, stoichiometry will help us find out how much ammonia is needed to produce a specific volume and concentration of nitric acid.

Molarity

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. It is crucial for understanding how much of a substance is present in a given volume of solution. In this question, the target concentration of 0.150 M for nitric acid will guide the calculations to determine the required amount of ammonia.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. Identifying the limiting reactant is essential in stoichiometric calculations, as it affects how much of the other reactants are needed. In this scenario, understanding which reactant limits the production of nitric acid will be key to solving the problem.