1°/ . 2 Al + 6 HCl → 2 AlCl3 + 3 H2
<span>k1 = n(Al) / 2 = 4,5 / 2 = 2,25 </span>
<span>k2 = n(HCl) / 6 = 11,5 / 6= 1,92 </span>
<span>k2 < k1 ==> HCl is the limiting reactant </span>
<span>6 mol of HCl ---> 2 mol of H2 </span>
<span>11,5 mol of HCl ---> 3,83 mol of H2 </span>
Answer:
No
Explanation:
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