compute theoretical yield (g)
~2g Al, 2g Cl2
i need the calculation work bcus i got diff from the ans which is 7.12 g.
moles Al = 2/26.98 = ??
moles Cl2 =2/70.9 = ??
moles Al x (2moles AlCl3/2 moles Al) = 0.0741 OR
moles Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.0188
Therefore, the limiting reagent is Cl2, moles AlCl3 formed is 0.0188 and that x molar mass of approximately 133 = 2.5 g AlCl3 as theoretical yield.posted by DrBob222
ME TOO. i was doing this question and got 7.4!! im so confusedposted by Anonymous