C%=\(\dfrac{150.4,9\%+100.20\%}{150+100}.100\%=10,94\%\)
Sửa chữa lỗi lầm:
2NaOH + H2SO4\(\rightarrow\)Na2SO4 + 2H2O
nH2SO4=\(\dfrac{150.4,6\%}{98}=0,075\left(mol\right)\)
nNaOH=\(\dfrac{100.20\%}{40}=0,5\left(mol\right)\)
Vì 0,075.2<0,5 nên NaOH dư 0,35 mol
Theo pTHH ta có:
nH2SO4=nNa2SO4=0,075(mol)
C% dd Na2SO4=\(\dfrac{0,075.142}{250}.100\%=4,26\%\)
C% dd NaOH=\(\dfrac{0,35.40}{250}.100\%=5,6\%\)