Question

Medal and Fan!!! find the limiting reagent, mass of Al_2Cl_6 that can be produced, and mass of excess that is remained in this equation when 2.70g Al and 4.05g cl2 are mixed

2 Al + 3Cl_2 = Al_2Cl_6

Answer 1

i found the limiting reagent which is the Cl_2 and the maximum Al2Cl6 that can be made is 5.08 g

Answer 2

Whats the Mm of Al2cl6?

Answer 3

266.66

Answer 4

\[5.08g Al_2Cl_2(\frac{1 mole Al_2Cl_6}{266.7g Al_2Cl_2})(\frac{2 mole Al}{1 mole Al_2Cl_6})(\frac{27.0g Al}{1 mole Al})=X g Al\]

Answer 5

2.70 g Al - X g Al = remaining reagents

Answer 6

thanks