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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 CI kas the smcallet vealue, it i be the a. What is the limiting reactant? Chitenne e limitng ieachant b. How many moles of the excess reactant are left?

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 CI kas the smcallet vealue, it i be the a. What is the limiting reactant? Chitenne e limitng ieachant b. How many moles of the excess reactant are left?

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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?
1.410 pmol CI, x mos,ly I50g SCla= 190.4 g of S,Cl2 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 (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 factor 100.00 oct, x mol Cl, (8) = 1,410 mol C 70.9 cla 200.0 48.x Imol Sa (R) = 0.7797 mol S, 256.5S 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. 1.410 mole Cl, available _ 1.808 mole Cl2 available 1 mol Se available Actual %3D Ratio 0.7797 mol Sg available To get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from the balanced chemical equation. Se mt 4 Clz (9 4 S2Cl2 ) tmoles Cl, 1 mole S Stoichiometric ratio = Step 5: Compare the actual ratio to the stoichlometric ratio The actual ratio tells us that we need 1.808 mole of Ca for every mole of Se. In the stoichiometric ratio, 4 moles of Clz is needed 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 mole ratio of the x limiting reactant and the product molar mass of the product Mass of the Product reactant 1.410 pmot Cl, x 1.350g 5,Cl; 1 1aol SCl = 190.4 g of S,Cla Now you know that 190.4 g of S:Cle is produced when 1.4 10 mol Cle reacts with an excess of Se. ( 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 a mole-to- mass calculation. The first step is to multiply the moles of chlorine by the mole ratio of sulfur to chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and the known in the denominator. 1.410 mol Cl, x 1 mol Sa 4 tal Cl. = 0,3525 mol Sa Now, to obtain the mass of sulfur needed, 0.3525 mol Se is muitiplied by the conversion factor that relates mass and moles, molar mass. 0.3525 mal Sa x E I eal Sa = 90.42 g Sy needest Knowing that 90 42 g Sa 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 Se available - 90,42 g Saa needed = 109.6 g Seg in excess.
Transcribed Image Text:1.410 pmol CI, x mos,ly I50g SCla= 190.4 g of S,Cl2 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 (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 factor 100.00 oct, x mol Cl, (8) = 1,410 mol C 70.9 cla 200.0 48.x Imol Sa (R) = 0.7797 mol S, 256.5S 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. 1.410 mole Cl, available _ 1.808 mole Cl2 available 1 mol Se available Actual %3D Ratio 0.7797 mol Sg available To get the stoichiometric ratio, divide the moles of chlorine to the moles of sulfur from the balanced chemical equation. Se mt 4 Clz (9 4 S2Cl2 ) tmoles Cl, 1 mole S Stoichiometric ratio = Step 5: Compare the actual ratio to the stoichlometric ratio The actual ratio tells us that we need 1.808 mole of Ca for every mole of Se. In the stoichiometric ratio, 4 moles of Clz is needed 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 mole ratio of the x limiting reactant and the product molar mass of the product Mass of the Product reactant 1.410 pmot Cl, x 1.350g 5,Cl; 1 1aol SCl = 190.4 g of S,Cla Now you know that 190.4 g of S:Cle is produced when 1.4 10 mol Cle reacts with an excess of Se. ( 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 a mole-to- mass calculation. The first step is to multiply the moles of chlorine by the mole ratio of sulfur to chlorine to obtain the number of moles of sulfur. Remember, the unknown is the numerator and the known in the denominator. 1.410 mol Cl, x 1 mol Sa 4 tal Cl. = 0,3525 mol Sa Now, to obtain the mass of sulfur needed, 0.3525 mol Se is muitiplied by the conversion factor that relates mass and moles, molar mass. 0.3525 mal Sa x E I eal Sa = 90.42 g Sy needest Knowing that 90 42 g Sa 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 Se available - 90,42 g Saa needed = 109.6 g Seg in excess.
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