Question

1mol／LAlCl3溶液75mL与2mol／LNaOH溶液120mL相混合,可得Al（OH）3沉淀多少克?

Answer 1

n(Al3+) = 1*0.075 = 0.075mol

n(OH-) = 2*0.12 = 0.24mol

Al3+ +3 OH- = Al(OH)3↓,

1 3 1

0.075 0.225 0.015

若Al3+完全沉淀则需要OH-的量为0.075*3 = 0.225 mol