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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 127

A volume of 18.72 mL of 0.1500 M K2Cr2O7 solution was required to titrate a sample of FeSO4 according to the equation in Problem 4.126. What is the mass of the sample?

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Hello everyone. So in this video we're being told that a chemical that produces the order of brown eggs is called hydrogen sulfide and it can be deodorized by reacting it with baking soda which has this compound or the chemical formula here. So here we can see that there's a reaction here that has our star regions of the baking soda as well as the hydrogen sulfide and yields these three different products. So first thing what I'm gonna do in this problem is to go ahead and balance out this reaction. So go ahead and rewrite this equation again, that's an A. H. C. 03 reacting with H two S. And that yields three different products, one being N. A two S. As well as c. 0. 2. And then H. 20. So I'm gonna go ahead and draw a line that separates our star materials from our products. And then I'll go ahead and keep a tally of all of our different um adam's here. Of course we have a we have a church C. Oh and us. Something on the right side of product side. That's an A. H. C. O. And us. So starting off just without any balancing or any adding any coefficients. We can see here that we have one sodium on the material side, we have three high regions, one carbon, three oxygen's and one sulfur. And for our product side we have two sodium to higher regions one carbon and three oxygen and one sulfur. So how can balance this is? We go ahead and add a confession to to our first story material to bounce out that sodium. So now we'll have to sodium we'll have four high regions, two carbons and six oxygen's. Now from my right side we also need to balance out our hydrogen carbon oxygen. How we can do this is by adding conditions to to our carbon dioxide as well as to for our liquid water. So now we can get two carbons and six oxygen's. So now everything is balanced. As you can see here in this tally. So this reaction right over here is going to be our balanced chemical reaction. I wish we were going to go ahead and use for a second step and our second step is our dimensional analysis. So again we're starting off with our 15 ml of our baking soda. So that's N. A. H. C. 03. We're gonna go ahead and convert our male leaders and to leaders. So for every 1000 ml we get one leader And then we'll go ahead and use the malaria T. or concentration of our baking soda as a conversion factor. So for every 0.893 moles baking soda we can get one liter of baking soda. Okay so we do so so far we can see that the male leaders will cancel as well as our leaders. Alright now using our balanced chemical reaction we can use the multiple ratio for our baking soda with our high hydrogen sulfide. So for every two moles of baking soda we get one mole, or we use one mole of H to us. And last see we can. We want to convert our mole units into grams, since it's asking how much of our hydrogen sulfide and grams can be derived arised with this amount of baking soda. So to get our moles into grams, we can use the molar mass of H two S, which is 34.1 g of H two S. For every one mole of H two S. So you can see now that the moles will go ahead and cancel the whole of baking soda as well as the moles of our hydrogen sulfide. So putting all these numerical values into the calculator, we get the value of 0.228 units being grams of H two S. And this right here is going to be my final answer for this problem. Thank you all so much for watching
Related Practice
Textbook Question
Iodine, I2, reacts with aqueous thiosulfate ion in neutral solution according to the balanced equation How many grams of I2 are present in a solution if 35.20 mL of 0.150 M Na2S2O3 solution is needed to titrate the I2 solution?
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Textbook Question
How many milliliters of 0.250 M Na2S2O3 solution is needed for complete reaction with 2.486 g of I2 according to the equation in Problem 4.124?
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Textbook Question
Dichromate ion, Cr2O7 2-, reacts with aqueous iron(II) ion in acidic solution according to the balanced equation What is the concentration of Fe2+ if 46.99 mL of 0.2004 M K2Cr2O7 is needed to titrate 50.00 mL of the Fe2+ solution?
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Textbook Question
What is the molar concentration of As(III) in a solution if 22.35 mL of 0.100 M KBrO3 is needed for complete reaction with 50.00 mL of the As(III) solution? The balanced equation is:
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Textbook Question
Standardized solutions of KBrO3 are frequently used in redox titrations. The necessary solution can be made by dissolving KBrO3 in water and then titrating it with an As(III) solution. What is the molar concentration of a KBrO3 solution if 28.55 mL of the solution is needed to titrate 1.550 g of As2O3? See Problem 4.128 for the balanced equation. (As2O3 dissolves in aqueous acid solution to yield H3AsO3: As2O3 + 3 H2OS 2 H3AsO3.)
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Textbook Question
The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to convert all the iron to Fe3+ and then titrated with Sn2+ to reduce the Fe3+ to Fe2+. The balanced equation is: What is the mass percent Fe in a 0.1875 g sample of ore if 13.28 mL of a 0.1015 M Sn2+ solution is needed to titrate the Fe3+?
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