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
