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Unbalanced equation 1. Given the following reaction: Si (s) + CI 2(g 35g SICI4 (1) ?g 56g When 56g of silicon are combined with 35g of chlorine gas in a reaction vessel: Smee CI has the Smallect value,it wi be the a. What is the limiting reactant? Chlenhe e lmiting ieachant b. How many moles of the excess reactant are left?

Unbalanced equation 1. Given the following reaction: Si (s) + CI 2(g 35g SICI4 (1) ?g 56g When 56g of silicon are combined with 35g of chlorine gas in a reaction vessel: Smee CI has the Smallect value,it wi be the a. What is the limiting reactant? Chlenhe e lmiting ieachant b. How many moles of the excess reactant are left?

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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SICI4 (1) Unbalanced equation 1. Given the following reaction: Si (s) + CI 2(g) 35g 56g ?g When 56g of silicon are combined with 35g of chlorine gas in a reaction vessel: Smee çI has the Smallect value,it wi be the a. What is the limiting reactant? Chlonhe lniting eachant b. How many moles of the excess reactant are left?
Transcribed Image Text:SICI4 (1) Unbalanced equation 1. Given the following reaction: Si (s) + CI 2(g) 35g 56g ?g When 56g of silicon are combined with 35g of chlorine gas in a reaction vessel: Smee çI has the Smallect value,it wi be the a. What is the limiting reactant? Chlonhe lniting eachant b. How many moles of the excess reactant are left?
Solution that should be followed. Thank youuu This kind of problem is an example of a limiting reactant problem since you are given the quantities of both the reactants and you are asked to calculate for the amount of the product. To solve limiting reactant problems, consider the following steps: Step 1: Write down the known and the unknown quantities in the problem. Given: mass sulfur = 200.0 g mass chlorine= 100 g Unknown: a) limiting reactant b.) 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 factor 100.00 ocl, x mel C (R) - 1.410 mo! C 70.a cla 1 mol Sp (a) - 0.7797 mol Su 256.5. 200,0 a8.x Step 4: Calculate the mole ratio of the reactants. To determine the actual ratio of moles, divide the available moles of chlorine by the available moles of sulfur which you calculated in Step 3. Actual 1.410 mole Clą available _ 1.808 mole Cl2 available Ratio %3D 0.7797 mol S, available 1 mol Se available To get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from the balanced chemical equation. Sa g+4 Clh w+ 4 S.Cbm tmoles Cl Stolchiometric ratio = mele S Step 5: Compare the actual ratio to the stoichiometric ratio The actual ratio tells us that we need 1.808 mole of Clafor every mole of Se. In the stoichiometric ratio, 4 moles of Clz is neeced for every mole of Sa. Since only1.808 moles of chlorine is actually available for every 1 mole of sulfur instead of the 4 mole of chlorine required by the balanced chemical equation then chlorine is the limiting reactant How to Get the Amount of Product Formed? Use the calculated amount of moles of the limiting reactant to determine the moles of product formed. 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 the reaction. To calculate: mole vale of the limiting molar mass of the product Mass of the Product mole ratio of the x limiting reactant X reactant and the nroduct 1.410 pro CI, x mgk5,x 15sag S,C 1 mol SCl, 190.4 g of S,Cla Now you know that 190.4 g of S.Cla is produced when 1.410 mol Cl, reacts with an excess e Se. ( 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 a mole-to- mass caiculation. The fist step is to multiply the moles of chlorine by the mole ratio of sultur to chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and the known in the denominator. 1 mel Se 0.3525 mol Sa 1.410 mol Cl, x 4 mot C. Now, to obtain the mass o sulfur needed, 0.3525 mol Sa is multiplied by the conversion factor that relates mass and moles, molar mass. 0.3525 mol S x **E 90.42 g S, needed - 90.42 g Sa needed I mal Sa Knowing that 90 42 g Ss is needed, you can calculate the amount of sulfur left unreacled when the reaction ends. Since 200.0 g of sulfur is available and only 90.42 g of sulphur is required, the excess mass is: 200.0 g Sea available - 90,42 g See needed = 109.6 g Seg in excess.
Transcribed Image Text:Solution that should be followed. Thank youuu This kind of problem is an example of a limiting reactant problem since you are given the quantities of both the reactants and you are asked to calculate for the amount of the product. To solve limiting reactant problems, consider the following steps: Step 1: Write down the known and the unknown quantities in the problem. Given: mass sulfur = 200.0 g mass chlorine= 100 g Unknown: a) limiting reactant b.) 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 factor 100.00 ocl, x mel C (R) - 1.410 mo! C 70.a cla 1 mol Sp (a) - 0.7797 mol Su 256.5. 200,0 a8.x Step 4: Calculate the mole ratio of the reactants. To determine the actual ratio of moles, divide the available moles of chlorine by the available moles of sulfur which you calculated in Step 3. Actual 1.410 mole Clą available _ 1.808 mole Cl2 available Ratio %3D 0.7797 mol S, available 1 mol Se available To get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from the balanced chemical equation. Sa g+4 Clh w+ 4 S.Cbm tmoles Cl Stolchiometric ratio = mele S Step 5: Compare the actual ratio to the stoichiometric ratio The actual ratio tells us that we need 1.808 mole of Clafor every mole of Se. In the stoichiometric ratio, 4 moles of Clz is neeced for every mole of Sa. Since only1.808 moles of chlorine is actually available for every 1 mole of sulfur instead of the 4 mole of chlorine required by the balanced chemical equation then chlorine is the limiting reactant How to Get the Amount of Product Formed? Use the calculated amount of moles of the limiting reactant to determine the moles of product formed. 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 the reaction. To calculate: mole vale of the limiting molar mass of the product Mass of the Product mole ratio of the x limiting reactant X reactant and the nroduct 1.410 pro CI, x mgk5,x 15sag S,C 1 mol SCl, 190.4 g of S,Cla Now you know that 190.4 g of S.Cla is produced when 1.410 mol Cl, reacts with an excess e Se. ( 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 a mole-to- mass caiculation. The fist step is to multiply the moles of chlorine by the mole ratio of sultur to chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and the known in the denominator. 1 mel Se 0.3525 mol Sa 1.410 mol Cl, x 4 mot C. Now, to obtain the mass o sulfur needed, 0.3525 mol Sa is multiplied by the conversion factor that relates mass and moles, molar mass. 0.3525 mol S x **E 90.42 g S, needed - 90.42 g Sa needed I mal Sa Knowing that 90 42 g Ss is needed, you can calculate the amount of sulfur left unreacled when the reaction ends. Since 200.0 g of sulfur is available and only 90.42 g of sulphur is required, the excess mass is: 200.0 g Sea available - 90,42 g See needed = 109.6 g Seg in excess.
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