1.410 pmol CI, x mos,ly I50g SCla= 190.4 g of S,Cl2Solution that should be followed. ThankyouuuThis kind of problem is an example of a limiting reactant problem since you are given thequantities of both the reactants and you are asked to calculate for the amount of the product. To solvelimiting reactant problems, consider the following steps:Step 1: Write down the known and the unknown quantities in the problem.Given: mass sulfur = 200.0 gmass chlorine= 100 gUnknown: a.) limiting reactantb.) mass of disulfur dichloride (S2Ch)Step 2: Balance the chemical equation.In the problem, the chemical equation is already balanced.Step 3: Convert mass of reactants to moles.Use the molar mass (inverse ) as a conversion factor100.00 oct, x mol Cl, (8) = 1,410 mol C70.9 cla200.0 48.x Imol Sa (R) = 0.7797 mol S,256.5SStep 4: Calculate the mole ratio of the reactants.To determine the actual ratio of moles, divide the available moles of chlorine by the available molesof sulfur which you calculated in Step 3.1.410 mole Cl, available _ 1.808 mole Cl2 available1 mol Se availableActual%3DRatio0.7797 mol Sg availableTo get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from thebalanced chemical equation.Se mt 4 Clz (9 4 S2Cl2 )tmoles Cl,1 mole SStoichiometric ratio =Step 5: Compare the actual ratio to the stoichlometric ratioThe actual ratio tells us that we need 1.808 mole of Ca for every mole of Se. In the stoichiometricratio, 4 moles of Clz is needed for every mole of Sa. Since only1.808 moles of chlorine is actuallyavailable for every 1 mole of sulfur instead of the 4 mole of chlorine required by the balanced chemicalequation then chlorine is the limiting reactantHow to Get the Amount of Product Formed?Use the calculated amount of moles of the limiting reactant to determine the moles of productformed. Then, convert the number of moles of product to its mass.Going back to the problem, we are asked of the mass of disulfur dichloride produced in thereaction. To calculate:mole vale ofthe limitingmole ratio of thex limiting reactantand the productmolar mass ofthe productMass of theProductreactant1.410 pmot Cl, x1.350g 5,Cl;1 1aol SCl= 190.4 g of S,ClaNow you know that 190.4 g of S:Cle is produced when 1.4 10 mol Cle reacts with an excess ofSe. ( Note: This is the theoretical yield)How to get the Excess Reactant?What about the reactant sulfur, which 13 ɔw is in excess? How much of it actually reacted?You can calculate the mass of sulfur needeu u react completely with 1.410 mol of chlorine using amole-to- mass calculation. The first step is to multiply the moles of chlorine by the mole ratio of sulfurto chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and theknown in the denominator.1.410 mol Cl, x1 mol Sa4 tal Cl.= 0,3525 mol SaNow, to obtain the mass of sulfur needed, 0.3525 mol Se is muitiplied by the conversion factorthat relates mass and moles, molar mass.0.3525 mal Sa x EI eal Sa= 90.42 g Sy needestKnowing that 90 42 g Sa is needed, you can calculate the amount of sulfur left unreacled whenthe reaction ends. Since 200.0 g of sulfur is available and only 90.42 g of sulphur is required, theexcess mass is:200.0 g Se available - 90,42 g Saa needed = 109.6 g Seg in excess. 2. Take the reaction: NH3 + 02 → NO + H2O. In an experiment, 3.25g of NH3 are aliowed toreact with 3.50 g of O2.a. Which reactant is limiting reagent?b. How many grams of NO are formed?c. How much of the excess reactant remains after the reaction?

Answered: Image /qna-images/answer/b2fdca87-25a6-4624-b6c6-47718264cd91.jpg

2. Take the reaction: NH3 + 02 → NO + H2O. In an experiment, 3.25g of NH3 are aliowed to react with 3.50 g of O2. a. Which reactant is limiting reagent? b. How many grams of NO are formed? c. How much of the excess reactant remains after the reaction?

2. Take the reaction: NH3 + 02 → NO + H2O. In an experiment, 3.25g of NH3 are aliowed to react with 3.50 g of O2. a. Which reactant is limiting reagent? b. How many grams of NO are formed? c. How much of the excess reactant remains after the reaction?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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