取纯铝4.05克放入0.5mol\lNaOH溶液500ml中,.取纯铝4.05克放入0.5mol\lNaOH溶液500ml中,生成的气体在标况下为多少升?在所得溶液中加入18.25%的盐酸80g,最终生成沉淀多少颗?

参照此题：
将0.1molAl投入3mol/LNaOH溶液100mL中，充分反应后再滴入1mol/L H2SO4溶液，若要得到全部沉淀，需要加入多少1mol/L H2S04溶液?

2Al + 2OH- + 6H2O ----> 2[Al(OH)4]- + 3H2
to completely react with 0.1 mol of Al, 0.1 mol OH- is required. There is still excess OH- left behind, of which the mol = 0.2 mol.
To neutralize 0.2 mol OH- requires 1 M H2SO4
V = 0.2/2 =0.1 L = 100 mL
Since the final precipitate is Al(OH)3
[Al(OH)4]- + H+ ----> Al(OH)3 + H2O
to completely convert [Al(OH)4]- to Al(OH)3 requires
V'=0.1/2 =0.05 L = 50 mL
so V(total) = V + V' = 150 ml of H2SO4

2Al+2NaOH+2H2O=2NaAlO2+3H2由方程式可以知道2mol的Al只能和2mol的NaOH反应纯铝4.05克的物质的量是：4.05/27=0.15mol0.5mol\lNaOH溶液500ml的物质的量是：0.5x0.5=0.25mol所以NaOH有剩余,用Al的质量计算2Al+2NaOH+2H...