How many grams of #AlCl_3# can be produced by reacting 80.0 grams of #Cl_2# with an unlimited supply of #Al#?

Your starting point here will be the balanced chemical equation for this synthesis reaction.

Aluminium metal, #"Al"#, will react with chlorine gas, #"Cl"_2#, to produce aluminium chloride, #"AlCl"_3#, according to the balanced chemical equation

#2"Al"_text((s]) + color(red)(3)"Cl"_text(2(g]) -> color(blue)(2)"AlCl"_text(3(s])#

Now, you know that aluminium is in excess, which implies that the reaction will completely consume the chlorine gas.

Notice that you have a #color(red)(3):color(blue)(2)# mole ratio between chlorine gas and aluminium chloride. This tells you that for every three moles of chlorine gas that take part in the reaction, you get two moles of aluminium chloride.

Use the molar mass of chlorine gas to determine how many moles you have in that #"80.0-g"# sample

#80.0 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "1.128 moles Cl"_2#

Next, use the aforementioned mole ratio to determine how many moles of aluminium chloride will be produced by this many moles of chlorine gas

#1.128color(red)(cancel(color(black)("moles Cl"_2))) * (color(blue)(2)color(white)(a)"moles AlCl"_3)/(color(red)(3)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7520 moles AlCl"_3#

Finally, to find the mass of aluminium chloride that would contain that many moles, use the compound's molar mass

#0.7520color(red)(cancel(color(black)("moles AlCl"_3))) * "133.34 g"/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = "100.27 g"#

Rounded to three **sig figs, the number of sig figs you have for the mass of chlorine gas, the answer will be

#m_(AlCl_3) = color(green)(|bar(ul(color(white)(a/a)"100 g"color(white)(a/a)|)))#