When 4.5 mol al react with 11.5 mol hcl, what is the limiting reactant, and how many moles of h2 can be formed?

1 answer:

5 0

1°/ . 2 Al + 6 HCl → 2 AlCl3 + 3 H2

k1 = n(Al) / 2 = 4,5 / 2 = 2,25 
k2 = n(HCl) / 6 = 11,5 / 6= 1,92 

k2 < k1 ==> HCl is the limiting reactant 

6 mol of HCl ---> 2 mol of H2 
11,5 mol of HCl ---> 3,83 mol of H2

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