Respuesta :

2Al + 6HCl = 2AlCl₃ + 3H₂

m(Al)=7.20 g
Vm = 22.4 L/mol
M(Al)=27.0 g/mol

n(Al)=m(Al)/M(Al)
n(H₂)=3n(Al)/2

V(H₂)=Vm n(H₂) = 3Vm n(Al)/2 = 3Vm m(Al)/{2M(Al)}

V(H₂)=3*22.4*7.20/{2*27.0}=8.96 L

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