Complete question is;
An excess of aluminum foil is allowed to react with 225 ml of an aqueous solution of HCl (d=1.088 g/ml) that contains 18% HCl by mass. what mass of H2(g) is produced?
Answer:
1.218 g
Explanation:
We are given d_HCL = 1.088 g/ml
Converting to g/L gives;
d = 1.088 × 1000 = 1088 g/L
Now, we are told the solution contains 18% HCl by mass.
Thus;
d = 1088 × 18%
d = 195.84 g/L
Now, molar mass of HCl is 36.46 g/mol
Thus, molar concentration is;
M = 195.84/36.46
M = 5.37 mol/L
We are told the volume of HCl used was 225 ml or 0.225 L
Thus, number of moles used is;
n = 5.37 × 0.225
n = 1.208
The balanced reaction equation of the aluminum foil and HCl is;
2Al + 3HCl = 2AlCl3 + 3H2
Number of moles of H2 produced will be; n1 = 1.208/2 = 0.604 moles
Molar mass of H2 is 2.016 g/mol
Thus, mass of H2 produced = 0.604 × 2.016 = 1.218 g
