a)
$Na_2O + H_2SO_4 \to Na_2SO_4 + H_2O$
b)
Theo PTHH :
$n_{Na_2SO_4} = n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$m_{dd\ sau\ pư} = 12,4 + 200 = 212,4(gam)$
$\Rightarrow C\%_{Na_2SO_4} = \dfrac{0,2.142}{212,4}.100\% = 13,37\%$
a)\(Na_2O+H_2SO_4\rightarrow Na_2SO+H_2O\)
0,2 → 0,2 →0,2
b)\(M_{dd}\) pư là:\(M_{Na_2O}+m_{H_2SO_4}\)
\(=12,4+200=212,4\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{0,2.142.100\%}{212,4}=13,4\left(\%\right)\)
a)
\(PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
b)
\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ \rightarrow n_{NaOH}=0,2.2=0,4\left(mol\right)\\ \rightarrow n_{Na_2SO_4}=142.0,2=28,4\left(g\right)\\ m_{ddNa_2SO_4}=m_{Na_2O}+m_{ddH_2SO_4}=12,4+200=212,4\left(g\right)\\ \rightarrow C\%_{ddNa_2SO_4}=\dfrac{28,4}{212,4}.100\approx13,371\%\)
