Chemistry msg for 78 mgSO 5.4 answer is NH3
Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so

Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.
The solution for this problem is:
C6H5NH3Cl is a strong salt: C6H5NH3+ + Cl- C6H5NH3+ + H2O <-----> C6H5NH2 + H3O+
K = Kw/ Kb= 1.0 x 10^-14 / 3.8 x 10^-10= 2.6 x 10^-5
=2.6 x 10^-5 = x^2 / 0.240-x
x = [H3O+] = 0.00251 M
pH = 2.60 is the concentration
The fist one on the fist page is a students walking home at the same pace because the line goes straight