191问答库 > 化工厂排放的废水中含有硫酸，为防止污染，现用2％的氢氧化钠来中和。完全反应后消耗了

化工厂排放的废水中含有硫酸，为防止污染，现用2％的氢氧化钠来中和。完全反应后消耗了

2025-03-17 02:52:59
推荐回答（2个）
回答1：

H2SO4+2NaOH=Na2SO4+2H2O
98 80
x 40g*2%
98/x=80/0.8g
x=0.98g
硫酸溶液中溶质的质量分数
0.98g/(138g-40g)*100%=1%

回答2：

40g2%的氢氧化钠溶液中氢氧化钠的质量为40g×2%=0.8g
则由
2NaOH+H2SO4====Na2SO4+H2O
80 98
0.8g 0.98g
而反应后溶液质量=反应前氢氧化钠溶液的质量+反应前硫酸溶液的质量
因此可得
反应前硫酸溶液的质量=138g-40g=98g
所以硫酸溶液的质量分数=0.98g/98g×100%=1%

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