CHEM 210 1st Edition Lecture 12Outline of Last Lecture I. Calculating Mass/PercentII. Empirical FormulaIII. IsomersOutline of Current Lecture IV. Using Balanced Equations for Word ProblemsV. Limiting ReactantsCurrent LectureIf you are presented with a word problem, you must first check if it is balanced before solving any of the questions it may ask you.For example,2Cu2S + 302 2Cu2O + 2SO2There are 4 moles of copper on the left side as well as the right side. There are 2 moles of sulfuron both sides of the equations as well, and finally there are 6 moles of oxygen on each side, so the equation is balanced.If the question from the problem asks you how many moles of O2 are required to roast 10.0 moles of copper (I) sulfide, you simply multiply by the mole ratio of oxygen and copper (I) sulfide, to get the answer.10.0 mol CuS2 X (3 mol O2) / (2 mol Cu2S) = 15.0 mol O2. If another problem asks you for the amount of grams of sulfur dioxide that forms when 10.0 molof copper (I) sulfide is roasted, you would multiply by the mole ratio of SO2 and Cu2S, and then multiply that by the molar mass of SO2 to end up with an answer in grams.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.10.0 mol Cu2S X (2 mol SO2) / (2 mol Cu2S) X (64. 07g SO2) / (mol SO2) = 641 g SO2. Finally, if the problem asks you how many kilograms of oxygen are required to form 2.86 kg of copper (I) oxide? You do the same steps to end up with an answer in kilograms.2.86 g Cu2O X (10^3 g Cu2O) / (1 kg Cu2O) X (1 mol Cu2O) / (143.10 g Cu2O) = 20.0 mol Cu2OThen,20.0 mol Cu2O X (3 mol O2) / (2 mol Cu2O) X (32.00 g O2) / (1 mol O2) X (1 kg O2) / (10^3 g O2) = 0.959 kg O2A Limiting Reactant is simply the reactant or reagent to be used up first in a chemical reaction. When faced with limiting reactants, your answer will depend on the limiting factors, rather thanthe entire
View Full Document
Unlocking...