n(HCl)=1*0.1=0.1mol
n(NaOH)=0.1mol
设Na xmol Al ymol
Na--------NaOH ------0.5H2 2Al + 2NaOH + 2H2O ====2NaAlO2 +3H2
1mol 1mol 0.5mol 2mol 2mol 3mol
x xmol b=0.5xmol ymol ymol a=3y/2
x=y+0.1
23x+27y=7.3
x=0.2mol y=0.1mol
原合金中Na,Al的质量:m(Na)=23*0.2=4.6g m(Al)=27*0.1=2.7g
反应过程中生成H2的物质的量:0.5*0.2+3*0.1/2=0.25mol