Given the reaction of 0.25 g of Al with HCl, how many liters of H_2H2 gas will be produced if the temperature is 40 C and the pressure is 1.15 atm?

2Al_((s)) + 6HCl_((aq)) -> 3H_(2(aq)) + 2AlCl_(3(aq))2Al(s)+6HCl(aq)→3H2(aq)+2AlCl3(aq)

1 Answer
anor277
Jan 18, 2017

Approx. 300*mL" of dihydrogen"300⋅mL of dihydrogen

Explanation:

You have the stoichiometric equation. We work out the molar equivalence of aluminum metal,

"Moles of Al"Moles of Al == (0.25*g)/(27.0*g*mol^-1)=9.26xx10^-3*mol0.25⋅g27.0⋅g⋅mol−1=9.26×10−3⋅mol

"Moles of dihydrogen"Moles of dihydrogen == 3/2xx9.26xx10^-3*mol=0.0139*mol32×9.26×10−3⋅mol=0.0139⋅mol

And V=(nRT)/PV=nRTP, where n=0.0139*moln=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

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