CHM 101 1st Edition Lecture 6Overview of lecture 5- Moles - Molar weight Lecture 6 I. Formula and Molecular WeightII. Limiting Reactants III. Theoretical YieldsCopper (I) sulfide reacts with oxygen to form copper (I) oxide and gaseous sulfur dioxide. How many grams of sulfur dioxide are produced when 305.0 g copper (I) sulfide react completely with oxygen?2Cu2S (s) + 3O2 (g)  2Cu2 (s) + 2SO2 (g)Cu2S 2(63.55) + 32.07 = 159.17 g/molSO232.07+2(16.00) = 64.07 g/mol305.0 g Cu2S x molCu2S x 2 mol SO2x 64.07 g SO2 159.17 g Cu2S 2 mol Cu2S mol SO2=122.8 g SO2Limiting Reactants - The reactant that is completely used up in a chemical reaction- The amount of product produced in a reaction is limited by the reactant that is present in the smallest stoichiometric amount- When two amounts of two reactants are given it is a limiting reaction problem- The reactant that produces less product is the limiting reactantThese 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.o Example Problem: How many grams of tetraphosphorusdecoxide (P4O10) can be produced by the reaction of 1.00 g phosphorus (P4) with 3.00 g oxygen? P4+ 5O2 P4O10- Determine the number of moles of product that would be formed from each reactant amount 1.00 g P4 x mol P4x 1 mol P4O10= 0.00807 mol P4O10 123.88 g P41 mol P43.00 g O2 xmol O2x 1 mol P4O10= 0.01875 mol P4O10 32.00 g O25 mol O2Use the limiting factor amount to finish the equation0.00807 mol P4O10x 283.88 g P4O10= 2.29 g P4O10Mol P4O10Theoretical Yields- Maximum amount of product that can be formed from the limiting reactant in a chemical reactionActual Yield - Actual amount of product formed during a chemical reactiono This amount is always less than the theoretical yieldPercent Yield- Relates the actual yield to the theoretical yield% yield = actual yield / theoretical yield x 100%When 100.0 kg sand (SiO2) are processed with carbon by the following reaction, 51.4 kg SiC are recovered. What is the % yield of this process?SiO2 (s) + 3 C (s) → SiC (s) + 2 CO (g)- Convert the kg to g- Then convert the grams to moles- Then covert it back to kg - Then use the % yield formulao 100.0kg SiO2 x 1000g SiO2 x 1 mol SiO2 x 1 molSiC x 40.10g SiC x 1kg SiC 1kg SiO260.09g SiO21 molSiC 1 molSiC 1000g SiC= 66.73kg SiC51.4 x 100% =