When 6.0 mol al react with 13 mol hcl, what is the limiting reactant, and how many moles of h2 can be formed? 2al + 6hcl → 2alcl3 + 3h2

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.

Hope this helps!!! :)

im not exactly sure but i think its b!