\(a,m_{NaOH\left(bđ\right)}=15\%.200=30\left(g\right)\\ Đặt:a=m_{NaOH\left(thêm\right)}\left(a>0\right)\left(g\right)\\ C\%_{ddNaOH\left(sau\right)}=17,07\%\\ \Leftrightarrow\dfrac{30+a}{200+a}.100\%=17,07\%\\ \Leftrightarrow a=5\left(g\right)\\ \Rightarrow m_{NaOH\left(thêm\right)}=5\left(g\right)\\ b,n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ m_{NaOH\left(ddY\right)}=0,5.2.40+30=70\left(g\right)\\ C\%_{ddY}=\dfrac{70}{200+31}.100\%\approx30,303\%\)