2Al+6HCl=2AlCl3+3H2
We calculate the theoretical mass from the reaction
For actual mass HCL 8.50g such actual mass Al is required:
, x=2.095
2.095 g is necessary for interaction with 8.50 g of hydrochloric acid
therefore, we find an excess of aluminum reagent
4.36-2.095=2.265
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