How do I find the last part for D [Review Topics]This question has multiple parts. Work all the parts to get the most points.Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine.2 Al(s) + 3 Cl₂(g) → 2 AlCl3 (8)If you begin with 2.70 g ofAl and 4.96 g ofCl₂,F2a which reactant is limiting?Cl₂Drag and drop your selection from the following list to complete the answer:AlAIC132.70 g Al x1. Calculate the amounts of each reactant.2. Compare the ratio of reactants amounts to the stoichiometric ratio defined by the equation.4.96 g Cl₂ x1 mol Al26.98 g0.100 mol Al0.0700 mol Cl₂Cl₂ is the limiting reactant.80F31 mol Cl₂70.90 g== 0.100 mol Al1.43 mol Al>1 mol Cl₂b What mass of AlCl3 can be produced?Show HintQ= 0.0700 mol Cl₂F42 mol Al3 mol Cl₂Cengage Learning | Cengage Technical Support0F5CCorrectF6«F7[References]DIIF8F91F10<EmailF11 bwhat mass of AlCl3 can be produced?Mass= 6.22g7.00 x 10-2 mol Cl₂ ×Calculate the mass of product based on the amount of the limiting reactant.2 mol AlCl33 mol Cla7.00 x 10-2 mol Cl₂ xX133.3 g1 mol AlCl3CWhat mass of the excess reactant remains when the reaction is completed?Mass=1.4419Submit[Review Topics]d Set up an amounts table for this problem.Equation2 Al(s)Initial amount (mol) 1.00x10^-1Change (mol)Show Hint2.70 g Al available - 1.26 g Al used = 1.44 g Al remainsDetermine the mass of the second reactant required in the reaction from the limiting reactant in a similar creagent is the difference between its starting mass and what was used in the reaction.Correct2 mol Al26.98 gX3 mol Cl₂= 1.26 g Al used1 mol Al6.22 g AIC13+ 3 Cl₂ (g)7.00x10^-2-7.00x10^-2Correct→ 2 AlCl3 (s)0[References]Cengage Learning | Cengage Technical Support

Given , Reaction : 2 AI (s) + 3 CI2 (g) → 2AICI3(s)

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How do I find the last part for D

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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Quantum Chemistry

Quantum chemistry deals with the properties of chemical systems and their reactions. The branch of chemistry which is more concerned with quantum chemistry is physical chemistry. It is a more sophisticated field of theoretical physics which is concerned towards the application of physical concepts towards various chemical aspects such as, formation of atoms, chemical structure descriptions, chemical bond strength, hybridization, conjugation etc. Quantum chemistry mainly studies the various states i.e., ground state, excited state, transition states of the individual atoms or molecules taking place during any chemical reaction. The laws of motion explained by various scientists, which are termed as “classical mechanics” were not acceptable in terms of electrons, protons, atoms, light etc. Thus, a new theory named “quantum mechanics” emerged to understand the basic framework of various atomic and subatomic particles.

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How do I find the last part for D

[Review Topics]
This question has multiple parts. Work all the parts to get the most points.
Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine.
2 Al(s) + 3 Cl₂(g) → 2 AlCl3 (8)
If you begin with 2.70 g of
Al and 4.96 g of
Cl₂,
F2
a which reactant is limiting?
Cl₂
Drag and drop your selection from the following list to complete the answer:
Al
AIC13
2.70 g Al x
1. Calculate the amounts of each reactant.
2. Compare the ratio of reactants amounts to the stoichiometric ratio defined by the equation.
4.96 g Cl₂ x
1 mol Al
26.98 g
0.100 mol Al
0.0700 mol Cl₂
Cl₂ is the limiting reactant.
80
F3
1 mol Cl₂
70.90 g
=
= 0.100 mol Al
1.43 mol Al
>
1 mol Cl₂
b What mass of AlCl3 can be produced?
Show Hint
Q
= 0.0700 mol Cl₂
F4
2 mol Al
3 mol Cl₂
Cengage Learning | Cengage Technical Support
0
F5
C
Correct
F6
«
F7
[References]
DII
F8
F9
1
F10
<
Email
F11

bwhat mass of AlCl3 can be produced?
Mass= 6.22
g
7.00 x 10-2 mol Cl₂ ×
Calculate the mass of product based on the amount of the limiting reactant.
2 mol AlCl3
3 mol Cla
7.00 x 10-2 mol Cl₂ x
X
133.3 g
1 mol AlCl3
CWhat mass of the excess reactant remains when the reaction is completed?
Mass=1.441
9
Submit
[Review Topics]
d Set up an amounts table for this problem.
Equation
2 Al(s)
Initial amount (mol) 1.00x10^-1
Change (mol)
Show Hint
2.70 g Al available - 1.26 g Al used = 1.44 g Al remains
Determine the mass of the second reactant required in the reaction from the limiting reactant in a similar c
reagent is the difference between its starting mass and what was used in the reaction.
Correct
2 mol Al
26.98 g
X
3 mol Cl₂
= 1.26 g Al used
1 mol Al
6.22 g AIC13
+ 3 Cl₂ (g)
7.00x10^-2
-7.00x10^-2
Correct
→ 2 AlCl3 (s)
0
[References]
Cengage Learning | Cengage Technical Support

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