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

$\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g$

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

5 0

3 years ago

Calculate the concentration of each species in a 0.240 m c6h5nh3cl solution.

amm1812

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

8 0

2 years ago

You can do Abc or 123 doesn't matter! Just give me the anwer pls

Olenka [21]

The fist one on the fist page is a students walking home at the same pace because the line goes straight

5 0

2 years ago