CHEM 111 1st Edition Lecture 6 Past Lecture Review:Avogadro ’s number (6.022 x 1023), Mass Percentage Calculation, Converting Moles and Mass, Converting Moles and Entities, Oxidation/Reduction, Neutralization, Combustion Reactions.Current Lecture: 2/06/2014Theoretical Yield:14.5 g Ga -----------> (Excess O2) of Ga2O3.Balance the Equation: 4 Ga + 3 O2 ----------> 2 Ga2O3Amount of Ga/molar mass of Ga14.5g / 69.72g/mol = .208 mol of Ga.208 mol x 2 mol Ga2O3 / 4 mol Ga = 0.104 mol Ga2O3**The 2 moles/4 moles comes from the underlined ratio in the balanced equation.Mass of Ga2O3: 187.4 g/mol x 0.104 mol = 19.5 g Ga2O3Limiting ReactantCalculate the mass of ammonia, NH3 product formed when 7.0 g of nitrogen, N2, react with 2.0 gof Hydrogen, H2.7.0 g N2 + 2.0 g H2 -----------------> NH3 g?1. Balance the Equation: N2 + 3 H2 ------------------> 2 NH32. Find out how many moles of nitrogen are in 7.0 g by finding the molar mass and diving 7.0 g by it.N2 = 7.0 g / 28.02 g/mol = .25 mol N2H2 = 2.0 g / 2.02 g/mol = 1 mol H2These 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.Thus, N2 is the limited reactant.3. Take each side of the equation and find out mole amounts.N2 + 3 H2 --------------> 2NH3.25 moles of limiting reactant, times the two from the balanced side = .5 mole NH3This leaves .25 moles of H2 left over.Actual Yield, Percent Yield- In most equations, not all of the product can be isolated.2P + 5 Cl2 -------------> 2PCl5What is the % yielded when 50 g of P reacts with 35 g of Cl2 and a chemist isolated 31.3 g of PCl52 P + 5 Cl2 --------------> 2 PCl550g 35g 31.3gP= 1.614 moles Cl=0.494 moles PCl= 0.150 molesP: 2 mol P ------> 2 mol PCl5 336.47 gCl2: 0.494 mol Cl2 ------> .198 mol PCl541.61 g (Theoretical) 75.22%Current Lecture Review:Limiting Reactants, How to calculate Theoretical yield, Actual Yield, Percentage