Question

How many grams of excess reagent are left over after the reaction of 4.36 g of Al and 8.50

g HCl in the following balanced chemical equation ?

Answer 1

2Al+6HCl=2AlCl₃+3H₂

We calculate the theoretical mass from the reaction

$m(Al)=n\times M=2\times27=54 g$

$m (HCl)=n\times M=6\times36.5=219 g$

For actual mass HCL 8.50g such actual mass Al is required:

$\frac{8.5}{219}=\frac{x}{54}$ , 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