an excess of aluminum foil is allowed to react with 225 ml of hcl that contains 18.0% hcl by mass what is mass of h2 is produced

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?

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

it takes on an electronical energy