Question

A sample of 9.00 grams of aluminum metal is added to an excess of hydrochloric acid. The volume of hydrogen gas produced at standard temperature and pressure is:

Answer 1

Answer:

The volumen of hydrogen (H) is 11.22 L

Explanation:

The balanced reaction between the aluminum (Al) and the hydrochloric acid (HCl) is:

2Al(s) + 6HCl(l) --> 2AlCl3(ac) + 3H2(g)

It is a reduction-oxidation reaction.

At the beginning we have 9.00 grams of Al what represents an certain amount of moles. Then, we know the molar mass of Al is 26.9815 g/mol, so the moles content in 9.00 g of Al are :

9.00 g Al * (1 mol Al / 26.9815 g Al)= 0.33356 mol Al

Now, we have to conisdered the molar relation between the Al and H2, according to the balanced reaction performed above. That is 2:3 (Al:H2)

Then,

0.33356 mol Al * (3 mol H2/  2mol Al) = 0.50034 mol H2

In this point, we can considered the H2 like a noble gas, because the question ask for standard temperature and pressure (0° C= 273.15K and 1 atm).

Let us remember the noble gases equation

PV=nRT

P: pressure

V: Volumen

n: number of moles

R: gases constant

T: temperature

For the volumen, the equation is:

V= (nRT/P)

Now: V= (0.50034 mol * 0.0821 (L*atm/K*mol) * 273.15 K)/ (1 atm)

         V= 11.22 L