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?
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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![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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78c34b25-9115-4867-ac70-85ef9ffb956c%2F4d2fcba7-8f1d-4f62-be0a-19f90eae93a2%2F7ipiozw_processed.jpeg&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78c34b25-9115-4867-ac70-85ef9ffb956c%2F4d2fcba7-8f1d-4f62-be0a-19f90eae93a2%2Fgj9bf4t_processed.jpeg&w=3840&q=75)
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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