(1)废水中氯化氢的质量为a
NaOH+HCl=NaCl+H 2O
40 36.5
4ga
40
4g =
36.5
a
解之得a=3.65g
废水中氯化氢的质量分数=
3.65g
1000g ×100%=0.365%
故答案为:0.365%;
(2)设处理100t这种废水,需要Ca(OH) 2的质量为x
Ca(OH) 2+2HCl=CaCl 2+2H 2O
7473
x100t×0.365%
73
100t×0.365% =
74
x
解得:x=0.37t
答:处理100t这种污水需要氢氧化钙的质量是0.37t.