Question

Giúp mình với ạ

Answer 1

\(n_{H_2SO_4}=0.06\cdot2=0.12\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(0.24...............0.12............0.12\)

\(V_{dd_{NaOH}}=\dfrac{0.24}{1}=0.24\left(l\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.12}{0.06+0.24}=0.4\left(M\right)\)

Answer 2

PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O(1)
              0,12    ->  0,24    ->   0,12                 (mol)
nH2SO4=Vdd.C(M)=0,06.2=0,12(mol)
-VddNaOH=n/C(M)=0,24/1=0,24(l)
VddSPỨ =VddH2SO4 + VddNaOH =0,06+0,24=0,3(l)
-C(M)ddA = C(M)Na2SO4=n/Vdd=0,12/0,3=0,4(M)