#M_r (Al) = 27.0#
mass given #= 0.150# g
#"number of moles" = "mass(g)"/"relative molecular mass (g per mol)"#
number of moles of #Al = "0.150 g"/"27.0 g/mol"# = 0.00556 mol
The reaction proceeds in a molar ratio of 2:2 (#Al:AlCl_3#)
Check we have 'enough' moles of #Cl_2# present in 1.00 g:
number of moles of #Cl_2 = "1.00 g"/"70.9 g/mol"# = 0.0141 mol
(molar ratio of #Al# to #Cl_2# is 2:3)
∴ #"0.00556 mol Al" xx "3 mol Cl"_2/"2 mol Al"# = 0.00833 mol of #Cl_2# needed
We have 0.0141 moles, so we have plenty left over!
So we can consider the #Cl_2# to be in excess and #Al# is the limiting reagent.
Therefore, if we have 0.00555 moles of #Al# then we will have 0.00556 moles of #AlCl_3#.
#"mass(g)" = "number of moles"xx "relative molecular mass (g.mol"^-1")"#
#"mass(g)" = "0.00556 mol" xx "133.3 g.mol"^-1#
#"mass = 0.741 g"#
