MasteringChemistry: RCT e1 CHEM 111 Review Essentials; Ch 1-8-Google Chrome Secu

Question

MasteringChemistry: RCT e1 CHEM 111 Review Essentials; Ch 1-8-Google Chrome Secure I https://session.masteringchemistry.com/myct/itemview?assignmentProblemID-95445140&offsets; next Constants | Periodic Table Aluminum metal reacts with chlorine gas to form solid aluminum chloride You are given 34.0 g of aluminum and 39.0 g of chlorine gas Part A What is the maximum mass of aluminum chloride that can be formed when reacting 34.0 g of aluminum with 39.0 g of chlorine? View Available Hint(s) Value Units Submit 11:16 AM pa ^ 4, 1/13/2018 Type here to search

Explanation / Answer

Molar mass of Al = 26.98 g/mol


mass(Al)= 34.0 g

use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(34.0 g)/(26.98 g/mol)
= 1.26 mol

Molar mass of Cl2 = 70.9 g/mol


mass(Cl2)= 39.0 g

use:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(39.0 g)/(70.9 g/mol)
= 0.5501 mol
Balanced chemical equation is:
2 Al + 3 Cl2 ---> 2 AlCl3


2 mol of Al reacts with 3 mol of Cl2
for 1.26 mol of Al, 1.89 mol of Cl2 is required
But we have 0.5501 mol of Cl2

so, Cl2 is limiting reagent
we will use Cl2 in further calculation


Molar mass of AlCl3,
MM = 1*MM(Al) + 3*MM(Cl)
= 1*26.98 + 3*35.45
= 133.33 g/mol

According to balanced equation
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.5501
= 0.3667 mol


use:
mass of AlCl3 = number of mol * molar mass
= 0.3667*1.333*10^2
= 48.89 g
Answer: 48.9 g

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