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MNaOH=40g/mol
n=m/M=16/40=0,4 moli NaOH
MAl=27g/mol
n=m/M=5,4/27=0,2 moli Al
NaOH este in exces => 0,4-0,2=0,2 moli NaOH in exces
2Al + 2NaOH + 2H2O => 2NaAlO2 + 3H2
b) 2 moli Al............................3 moli H2
0,2 moli Al............................x=3*0,2/2=0,3 moli H2
c) 2NaOH + H2SO4 => Na2SO4 + 2H2O
2*40g NaOH......................98g H2SO4
16g NaOH..........................x=19,6g H2SO4
C=md/ms*100=>ms=md*100/C=1960/49=40g sol H2SO4
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