Skip to main content
Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 144

An unknown metal (M) was found not to react with either water or steam, but its reactivity with aqueous acid was not investigated. When a 1.000 g sample of the metal was burned in oxygen and the resulting metal oxide converted to a metal sulfide, 1.504 g of sulfide was obtained. What is the identity of the metal?

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Video transcript

Welcome back everyone. The reactivity of a metal M was studied and it was recorded that it does not react with water or steam but can displace protons from acids to produce hydrogen gas. A mass of five or 1.500 g of the metal was reacted with oxygen to produce a metal oxide. And then the metal oxide was further reacted to form a metal sulfide. If 1.905g of the metal sulfide was produced, determined the identity of the metal. We want to recall our activity series table in our textbooks. And specifically, we want to recall this series of metals that displace hydrogen from assets. So we would recall the medal specifically beginning with cadmium since its highest up in our activity series which is corresponding to the cadmium two plus canyon below cadmium. On our activity series, we have our metal Cobalt which forms the cobalt two plus caddy. On below Cobalt, we have nickel forming the nickel to plus catalon and 10, we have which forms the S N two plus caddy on and then we have led which is the most bottom of our activity series forming the lead two plus catalon. And so all of these metals can displace hydrogen from acids. Now, because we need to determine the identity of our metal. We're going to make a general formula for our metal sulfide represented as M S. So we need to focus on first using the info from the pumped finding our mass of R S being our sulfide. So for the mass of our sulfide S, We would begin with taking the amount of the metal sulfide produced given in the prompt as 1.905g subtracted from the mass of the metal mentioned in the prompt as 1.500 g. And so this difference gives us our mass of our sulfide equal 2.405g. Now, with this mass of our sulfide in our metal sulfide, we want to find moles of our sulfide. And so beginning with our mass 0.405 g, we're going to multiply by the molar mass of sulfur in which on our product tables, we see is equivalent to 32.66 g of sulfur equivalent to one mole of sulfur. So canceling out grams, we're left with moles and this is going to result in a in our moles of our sulfide equal to 0. moles of our sulfide. And now that we have most of our sulfide, we can use it in a ratio to find moles of our metal. And so beginning with our moles of sulfide, 0.1263 moles of our sulfide. We're going to multiply by the molar ratio To go from moles of our sulfide, two moles of our metal. And in our, in our chemical formula for our metal sulfide, we should have a 1-1 ratio. So one mole of our sulfide for one mole of our metal using this ratio and calculating things into our calculator. We find that our, most of our metal are 0.1263 moles of our metal. And now we need to find the mass so that we can identify its identity. So we want to recall that for the molar mass of our mettle, we would take crams over moles. And so we know that our mass of our metal according to the prompt is 1.500 g. And our moles, we just calculated to be 0. moles. And so we have grams per mole in which we will have our molar mass equal to a value of 118. g. Sorry, this should be grams per mole. Now, with the smaller mass, we want to refer to our periodic table. So on our periodic table, we would see that this smaller mass corresponds to the mass of tin in which we have a mass of about 118.71 g per mole. And so we would therefore have identified our identity, our identity of the metal being tin. And so what's highlighted in yellow and actually, we shouldn't highlight the mass, but what's highlighted in yellow for the name of our metal being tin would be our final answer to complete this example corresponding to choice c in the multiple choice. I hope that everything I went through was clear. If you have any questions, please leave them down below and I'll see everyone in the next video.
Related Practice
Textbook Question

(b) If Ksp = 1.1 * 10-12 for Ag2CrO4, what are the molar concentrations of Ag+ and CrO4 2-in solution?

520
views
Textbook Question
A 100.0 mL solution containing aqueous HCl and HBr was titrated with 0.1235 M NaOH. The volume of base required to neutralize the acid was 47.14 mL. Aqueous AgNO3 was then added to precipitate the Cl-and Br-ions as AgCl and AgBr. The mass of the silver halides obtained was 0.9974 g. What are the molarities of the HCl and HBr in the original solution?
794
views
1
comments
Textbook Question
A sample of metal (M) reacted with both steam and aqueous HCl to release H2 but did not react with water at room tem-perature. When 1.000 g of the metal was burned in oxygen, it formed 1.890 g of a metal oxide, M2O3. What is the iden-tity of the metal?
600
views
Textbook Question
A mixture of acetic acid (CH3CO2H; monoprotic) and oxalic acid (H2C2O4; diprotic) requires 27.15 mL of 0.100 M NaOH to neutralize it. When an identical amount of the mixture is titrated, 15.05 mL of 0.0247 M KMnO4 is needed for complete reaction. What is the mass percent of each acid in the mixture? (Acetic acid does not react with MnO4 equation for the reaction of oxalic acid with MnO4 given in Problem 4.133.)
1110
views
Textbook Question
Iron content in ores can be determined by a redox procedure in which the sample is first reduced with Sn2+, as in Problem 4.130, and then titrated with KMnO4 to oxidize the Fe2+ to Fe3+. The balanced equation is What is the mass percent Fe in a 2.368 g sample if 48.39 mL of a 0.1116 M KMnO4 solution is needed to titrate the Fe3 + ?
306
views
Textbook Question
A mixture of FeCl2 and NaCl is dissolved in water, and addi-tion of aqueous silver nitrate then yields 7.0149 g of a pre-cipitate. When an identical amount of the mixture is titrated with MnO4 -, 14.28 mL of 0.198 M KMnO4 is needed for complete reaction. What are the mass percents of the two compounds in the mixture? (Na+ and Cl-do not react with MnO4 -. The equation for the reaction of Fe2+ with MnO4 was given in Problem 4.146.)
891
views