\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\\ m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH: Na2CO3 + H2SO4 ---> Na2SO4 + CO2 + H2O
LTL: 0,1 < 0,2 => H2SO4 dư
Theo pthh: \(n_{Na_2SO_4}=n_{H_2SO_4\left(pư\right)}=n_{CO_2}=0,1\left(mol\right)\)
\(m_{dd}=10,6+200-0,1.44=206,2\left(g\right)\\ C\%_{H_2SO_4}=\dfrac{98.0,1}{206,2}.100\%=4,75\%\\ C\%_{Na_2SO_4}=\dfrac{0,1.142}{206,2}.100\%=6,88\%\)
\(m_{\text{dd}H_2SO_4}=200.9,8\%=19,6g\)
\(m_{\text{dd}}=19,6+10,6=30,2\left(g\right)\)
\(\Rightarrow C\%=\dfrac{10,6}{30,2}.100\%=35\%\)
