Question

6. The following questions (parts A, B and C) are unrelated to each other and are based on the following reaction. 2Al(s) + 3Cl2(8) ► 2A1CH() Al Determine the limiting reactant when 56.2 grams of Al (molar mass 27.0 g/mol) are reacted with 198 grams of Cl2 (molar mass = 70.9 g/mol). 56.28 2mol = 4.16 MOLAI (27.08 not correct method 198 get xml) . 5.59 MOL CIL Al IS LIMITING REACTANT X B) If 5.00 moles of Cl, are reacted with an excess of Al, how many grams of AICI: (molar mass - 133 g/mol) are fomed? Soomolci2 x 1339 = 1665 g AlCl3 mol D need unito identifiers C) If 1.8 moles of Al are reacted with 8.6 moles of Cly, how many moles of Cl2 are left unreacted? (1.8 x 3/2 = 2.7mol 8.6 molts C12 -2.7 MOL (= 5.ama

question has 3 parts please help. due tonight

Answer 1

2Al + 3Cl2 ----------- 2AlCl3 is the reaction

Part A

2Al + 3Cl2 -----------> 2AlCl3

56.2g/27g/mol 198g/70.9g/mol 0

=2.081 =2.793 0 initial moles

to calculate the limiting reagent

find the ratio of given moles to reuired moles

2.081/2= 1.04 2.793/3 =0.93

As the ratio of Cl2 is less , CL2 IS THE LIMITING REAGENT.

Part B)

2Al + 3Cl2 -----------> 2AlCl3

from this equation

3 moles of Cl2 reacts to  form 2 moles of AlCl3

If 5 moles of Cl2reacts = 5mol Cl2 x 2 mol AlCl3 /3 molCl2=3.33mol AlCL3

mass of ALCl3 = moles x molar mass

= 3.33mol x 133g/mol =443.33 g

Part C

2Al + 3Cl2 -----------> 2AlCl3

1.8 8.6 0 inital moles

0 8.6-[1.8x3/2]=5.9 1.8 after reaction

Thus Cl2 left = 5.9 moles