LG Week 4 Answer Key (docx) - CliffsNotes (2024)

NOTE: I wrote down the rounded values for intermediate steps since its easier to type but I used the unrounded versions and so should you! Also if your answers are slightly off (last digit usually), might be because differences in periodic tables used 1). NH3and N2H4 both consist of nitrogen and hydrogen.Following the law of multiple proportions, for a fixed mass of nitrogen, their masses of hydrogen have what ratio? You need to fix the N's. Looking at the N:H ratios, they are 1:3 and 2:4 for in the two compounds. We can use that 1:3 is the same as 2:6 (to have 2's for the N's in both) so then we can directly compare the H's which have ratio have 6:4 or 3:2 for NH3to N2H4 (2:6 and 2:4 so H's are 6:4 if you want you can assume some masses ex assume both compounds have 28.04 g of N and do it out that way by getting to grams of H in both but you will get the same result). 2). Name or give the name of each of the following compounds: KCNPotassium Cyanide HBrO2 Bromous Acid AlH3 Aluminum Hydride S2O10 Disulfur Decoxide P2I4 Diphosphorus Tetraiodide Hg(NO3)2ᐧ2H2 OMercury(II) Nitrate Dihydrate HBr (g)Hydrogen Bromide (because its gas its not the acid) Sn(C2H3O2)2 Tin(II) Acetate C8H18 Octane Ba(OH)2ᐧ8H2 OBarium Hydroxide Octahydrate Au2S3 Gold (III) sulfide C7H16 Heptane OF2 Oxygen difluoride NH4C2H3O2 Ammonium acetate HBrO3 Bromic acid Cu2 OCopper (I) oxide HF(aq)Hydrofluoric acid

PbSO3 Lead (II) sulfite Na2HPO4 Sodium hydrogen phosphate CrI3 Chromium (III) iodide 3). Equal masses of Fe and H2 O are combined and the following reaction takes place: 3Fe(s) + 4H2O(l) → Fe3O4(s) + 4H2 (g) a.If the percent yield of this reaction is 81.25% due to an incomplete reaction and the actual yield of Fe3O4 is 70.46g, which reactant was present in excess and how many grams of that reactant will remain? (I use MM as shorthand for molar mass in g/mol) If actual yield is 70.46 g , theoretical is 70.46g/.8125 = 86.72 g. Then convert to mols Fe3O4 by dividing by molar mass and get .3745 mols product as the THEORETICAL YIELD (used later). To determine which reactant is limiting we can assume 100g of each since we know they were combined in equal mass. For water if we do this we can see that 100/MM water = roughly 5.55 mols water then divided by 4 (4 mols water used = 1 mol Fe3O4 formed) is roughly 1.39 mols product. On the other hand with iron its 100/MM iron = roughly 1.79 mols Fe and then divided by 3 = less than 1.39 mols product (3 mols Fe used = 1 mol Fe3O4 formed from coefficients) so the Fe is limiting (made less product). Now to find how much is left we use the theoretical yield based on the limiting reagent to figure out how much water we started with but the actual yield to figure out how much was used since the rxn was incomplete (if the reaction was complete then the theoretical yield would work both ways but since it stopped part way through that means there were some of BOTH reactants left over including the limiting one which normally does not happen). First find how much we started with: .3745 mols Fe3O4*3 mols Fe/mol Fe3O4 * MM Fe = 62.75 g Fe initially (know equal masses at start so also 62.75 g water!!!!) Have to turn to Iron since its limiting, cannot go to water directly as limiting determines both, if we went to water we would get a smaller initial amount then there actually was because water was excess meaning even if the reaction went to 100% completion there would be some leftover while the conversion from product to reactants is only based on how much is used. How much used = 70.46g (actual yield) /MM Fe3O4* 4 mols water/mol Fe3O4 * MM water = 21.93 g water reacted, take the difference 62.75 initial mass - 21.93 reacted mass, get 40.82 g remaining (4 SFs) We use the actual yield to determine the amount remaining because we are assuming the yield was lowered due to the reaction not being complete not due to human error so the amount

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