Question

200 de grame de hidroxid de sodiu de concentratie 20% reactioneaza cu 200 grame soluție de acid sulfuric de concentratie 49%

Answer 1

Notez inmultirile cu * si fractiile cu / 

2NaOH+H2SO4-->Na2SO4+2H2O 
MNaOH=23+16+1=40g/mol 
MH2SO4=2*1+32+4*16 = 98g/mol 
c1=md1*100/ms1=>md1=c1*ms1/100 =20*200/100=40 g NaOH 
c2=md2*100/ms2=>md2=c2*ms2/100=49*200/100=98 g H2SO4 
n NaOH=m/M=40g/40g/mol=1 mol 
n H2SO4 =m/M=98g/98g/mol = 1 mol  
Practic : 1 mol NaOH :1 mol H2SO4
Teoretic : 2 moli NaOH : 1 mol H2SO4  
m H2SO4  in exces = 49 g 

1 mol                             x            y 
2NaOh+H2SO4-->Na2SO4 +2H2O   
2 moli                       1 mol         2 moli 

1/2=x/1=>x=1*1/2=0.5 moli Na2SO4 
1/2=y/2=>y=2*1/2=1mol H2SO4 

nH2O = m/M=> m = n*M = 1*18 = 18g 

m apa final = mH2O sol NaOH+ mH2O sol H2SO4 +mH2O=200g-40g+200g-98g+18 g = 160+`102+18=280 g