$①升高\ $
$②2NaOH+Cl_{2}\xlongequal[\\]{\\}NaClO+NaCl+H_{2}O\ $
$③NaCl\ $
$④解：设生成氢气的质量为x，氢氧化钠的质量为y。\ $
$2NaCl+2H_{2}O\xlongequal[\\]{通电}Cl_{2}↑+H_{2}↑+2NaOH$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~71 ~~~~~~~~~2 ~~~~~~~~~80$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~7.1g~~~~~~ x~~~~~~~~~~ y\ $
$\frac {71}{2}=\frac {7.1g}{x}，\ $
$解得x=0.2g。\ $
$\frac{71}{80}=\frac {7.1g}{y}，\ $
$解得y=8g。\ $
$反应后所得溶液中溶质质量分数：$
$\frac {8g}{100g-7.1g-0.2g}×100\%≈8.6\%。\ $
$答：生成氢气的质量为0.2g，反应后溶液中NaOH的质量分数是8.6\%。\ $
$⑤HCl(或氯化氢)$
