Given the reaction of 0.25 g of Al with HCl, how many liters of #H_2# gas will be produced if the temperature is 40 C and the pressure is 1.15 atm?

Question

#2Al_((s)) + 6HCl_((aq)) -> 3H_(2(aq)) + 2AlCl_(3(aq))#

Wyatt Atkins · Accepted Answer

Approx. #300*mL" of dihydrogen"#

You possess the stoichiometric equation. We determine the aluminum metal's molar equivalency,

#"Moles of Al"# #=# #(0.25*g)/(27.0*g*mol^-1)=9.26xx10^-3*mol#

#"Moles of dihydrogen"# #=# #3/2xx9.26xx10^-3*mol=0.0139*mol#

And #V=(nRT)/P#, where #n=0.0139*mol#, so............

#(0.0139*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx313*cancelK)/(1.15*cancel(atm))#

#=??L#

Peyton Adams · Answer

undefined To find the volume of H2 gas produced, we first need to calculate the moles of H2 gas produced from the reaction of 0.25 g of Al with HCl using stoichiometry. Then, we can use the ideal gas law to find the volume of the gas at the given temperature and pressure.

1. Calculate moles of Al:
   $ moles_{Al} = \frac{mass_{Al}}{molar\,mass_{Al}} $

2. Use stoichiometry to find moles of H2 produced:
   The balanced chemical equation is:
   $ 2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 $
   From the equation, 2 moles of Al produce 3 moles of H2.
   $ moles_{H_2} = \frac{moles_{Al}}{2} \times 3 $

3. Use the ideal gas law to find the volume of H2 gas:
   $ PV = nRT $
   Where:
   $ P = 1.15 \, atm $ (pressure)
   $ V $ (volume) is what we're solving for
   $ n $ (moles) is the calculated moles of H2
   $ R = 0.0821 \, \frac{atm \cdot L}{mol \cdot K} $ (ideal gas constant)
   $ T = 40 + 273.15 = 313.15 \, K $ (temperature in Kelvin)
   
   $ V = \frac{nRT}{P} $

Let's calculate:

1. $ moles_{Al} = \frac{0.25 \, g}{26.98 \, g/mol} = 0.00927 \, mol $
2. $ moles_{H_2} = \frac{0.00927}{2} \times 3 = 0.0139 \, mol $
3. $ V = \frac{0.0139 \, mol \times 0.0821 \, \frac{atm \cdot L}{mol \cdot K} \times 313.15 \, K}{1.15 \, atm} = 0.298 \, L $

Therefore, approximately 0.298 liters of H2 gas will be produced.