SICI4 (1)Unbalanced equation1. Given the following reaction: Si (s) + CI 2(g)35g56g?gWhen 56g of silicon are combined with 35g of chlorine gas in a reaction vessel:Smee çI has the Smallect value,it wi be thea. What is the limiting reactant? Chlonhe lniting eachantb. How many moles of the excess reactant are left? Solution that should be followed. Thank youuuThis 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 (SzCh)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 ocl, x mel C (R) - 1.410 mo! C70.a cla1 mol Sp (a) - 0.7797 mol Su256.5.200,0 a8.xStep 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.Actual 1.410 mole Clą available _ 1.808 mole Cl2 availableRatio%3D0.7797 mol S, available1 mol Se availableTo get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from thebalanced chemical equation.Sa g+4 Clh w+ 4 S.Cbmtmoles ClStolchiometric ratio = mele SStep 5: Compare the actual ratio to the stoichiometric ratioThe actual ratio tells us that we need 1.808 mole of Clafor every mole of Se. In the stoichiometricratio, 4 moles of Clz is neeced 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 limitingmolar mass ofthe productMass of theProductmole ratio of thex limiting reactant Xreactantand the nroduct1.410 pro CI, x mgk5,x 15sag S,C1 mol SCl,190.4 g of S,ClaNow you know that 190.4 g of S.Cla is produced when 1.410 mol Cl, reacts with an excess eSe. ( Note: This is the theoretical yield)How to get the Excess Reactart?What about the reactant sulfur, which 13 Jw is in excess? How much of it actually reacted?You can calcuiate the mass of sulfur needeu u Ieact completely with 1.410 mol of chlorine using amole-to- mass caiculation. The fist step is to multiply the moles of chlorine by the mole ratio of sulturto chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and theknown in the denominator.1 mel Se 0.3525 mol Sa1.410 mol Cl, x4 mot C.Now, to obtain the mass o sulfur needed, 0.3525 mol Sa is multiplied by the conversion factorthat relates mass and moles, molar mass.0.3525 mol S x **E 90.42 g S, needed- 90.42 g Sa neededI mal SaKnowing that 90 42 g Ss 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 Sea available - 90,42 g See needed = 109.6 g Seg in excess.

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