a)GS $H2SO4$ pứ x mol
=>mdd Y=7,4+348-2x=355,4-2x gam
Kiềm dư=>ktủa là $Mg(OH)2$ 0,1 mol
Sau khi cho td vs kiềm mddZ=60+355,4-2x-5,8=409,6-2x gam
=>2x=0,6=>x=0,3 mol
Gọi n$Na$=y n$Al$=z mol
=>n$H2SO4$=0,5y+1,5z+0,1=0,3
mhh=7,4=0,1.24+23y+27z
=>y=z=0,1
=>%m$Mg$=32,43% %m$Al$=36,49% %m$Na$=31,08%
b)m$H2SO4$=0,3.98=29,4g=>C% dd $H2SO4$=8,45%
c)n$NaOH$=0,66 mol
$MgSO4$+2NaOH=>$Mg(OH)2$+$Na2SO4$
$Na2SO4$(Y)=0,1:2=0,05 mol
$Al2(SO4)3$+6$NaOH$=>2$Al(OH)3$+3$Na2SO4$
$Al(OH)3$+$NaOH$=>$NaAlO2$+2$H2O$
n$NaOH$ pứ=0,1.2+0,05.6+0,1=0.6 mol
=>n$NaOH$ dư=0,06 mol
Vậy Z chứa $NaOH$ dư Na2SO4 và $NaAlO2$
C%dd $NaOH$ dư=0,06.40/409.100%=0,587%
C%dd$NaAlO2$=0,1.82/409.100%=2%
C%dd$Na2SO4$=(0,05+0,1+0,05.3).142/409.100%=10,42%