\(4/a.n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ Na_2O+H_2O\xrightarrow[]{}2NaOH\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ C_{MNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ n_{H_2SO_4}=n_{NaOH}=0,5mol\\ m_{H_2SO_4}=0,5.98=49\left(g\right)\\ V_{H_2SO_4}=\dfrac{49}{1,14}=42,98\left(ml\right)\)
\(5/a.n_{CO_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\\ n_{NaOH}=\dfrac{6,4}{40}=0,16\left(mol\right)\\ 2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ \Rightarrow\dfrac{0,16}{2}>\dfrac{0,07}{1}\Rightarrow NaOH.dư\\n_{Na_2CO_3}=n_{CO_2}=0,07mol\\ m_{Na_2CO_3}=0,07.106=7,42\left(g\right)\\ b.n_{NaOH\left(dư\right)}=0,16-0,07.2=0,02\left(mol\right)\\ m_{NaOH\left(dư\right)}=0,02.40=0,8\left(g\right)\)
