Hey, i would like to know how many moles of AlCl(3) were produced, and which substance is the limiting reactant?
(The following reaction was carried out using 1 mole of Al and 1 mole of Cl(2)
2Al + 3Cl(2) ------------> 2AlCl(3)
tanks
2007-03-20 05:02:02 · 3 answers · asked by Anonymous in Education & Reference ➔ Homework Help
1 mol Al x (2 mol AlCl3 / 2 mol Al) = 1 mol AlCl3
1 mol Cl2 x (2 mol AlCl3 / 3 mol Cl2) = 0.666 mol AlCl3 [Limiting]
Limiting Reactant -- will produce the least amount of product
2007-03-20 05:15:17 · answer #1 · answered by Pseudosophy 3 · 0⤊ 1⤋
For every two molecules of Al used, 3 molecules of Cl(2) are used. Thus, since you have equal amounts of the two, the Cl(2) gets used up faster. Then, find the number of moles of Al that was used (2/3 mol) and you'll get your answer for how much AlCl(3) was created.
2007-03-20 05:13:59 · answer #2 · answered by Ted G 2 · 0⤊ 1⤋
Sorry to waste your time checking this.. But I had to say.. That default pic is WAY to funny ! Good luck with finding the answer and sorry i didnt have it.
2007-03-20 05:16:10 · answer #3 · answered by dralls4lyf 2 · 1⤊ 0⤋