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(化学)一道化学选择题,帮帮忙啊!^_^

2025年04月03日 20:35

有3个网友回答

网友（1）：

将2.7gAl与100mL2mol/LH2SO4溶液充分反应后，此时硫酸过量，剩余0。1mol氢离子，恰好和加入的25mL4mol/LNaOH溶液完全反应。此时溶液中存在的溶质是AL2(SO4)3和Na2SO4，但是AL2(SO4)3是强酸弱碱盐水解成酸性，所以溶液pH<7，答案A错误。B正确.C选项加NaHCO3，是强碱弱酸盐，遇到AL2(SO4)3发生双水解反应，生成和CO2气体。所以C错误。选项D正确会产生Al(OH)3沉淀

网友（2）：

a

网友（3）：

0.1mol铝和0.4mol氢离子反应，生成0.1mol铝离子和0.3/2mol氢气，剩下0.1mol氢离子。再加上0.1mol氢氧根离子.
应该是中性的，加上铝离子水解，显偏酸性——A错
铝离子和碳酸氢钠双水解，沉淀冒泡——C不对
所以应该是BD正确

方程式：（1）2Al+3H2SO4=Al2(SO4)3+3H2
(2)2NaOH+H2SO4=Na2SO4+2H2O
气体和沉淀符号，麻烦您自己写

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