How many molecules of Cl2 must react to produce 24.6g of AlCl3?

1 Answer
Apr 13, 2018

#bb(2Al + 3Cl_2 rarr 2AlCl_3#

According to the equation, #3# moles of #Cl_2# react to give, #2# moles of #AlCl_3#
So, #1# mole of #AlCl_3# requires #3/2# moles of #Cl_2#

Moles of #AlCl_3# = #"given mass"/"molar mass"#

#=> 24.6/133.3 = 0.185#

Therefore we need, #3/2xx0.185# moles of #Cl_2 = 2.7 #moles.

Also,

Moles of #Cl_2# = #"given number of particles"/"avagrados number"#

#=> "number of molecules required" = 2.7 xx 6.022 xx 10^23 = 1.62 xx 10^24#