Question

1.将0.1mol Al投入到100mL 3mol/L NaOH溶液中,充分反应后再滴加1mol/L硫酸溶液120mL,

Answer 1

1.

n(NaOH)=3X0.1=0.3mol

n(H2SO4)=1x0.12=0.12mol

2Al+2NaOH+2H2O=NaAlO2+3H2有方程式可以判断,NaOH过量,用Al的量计算

2Al+2NaOH+2H2O=NaAlO2+3H2

2.2

0.1..x

x=0.1mol

剩余n(NaOH)=0.3mol-0.1=0.2mol

2NaOH+H2SO4=Na2SO4+2H20有方程式可以判断,H2SO4过量,

所以H2SO4 有剩余,剩余n(H2SO4)=0.12-0.1=0.02mol

H2SO4+2NaAlO2+2H2O=2AL(OH)3+Na2SO4

所以有沉淀,NaAlO2有剩余,PH>7

选A

2.设沉淀质量是m,

Al3+ + 3OH-=Al(OH)3,Al(OH)3+ OH-=AlO2 - + 2H2O

那么Al(OH)3中Al质量是：mx27/78=9m/26

所以：AlO2 -中Al质量是:m-9m/26=15m/26

Al3+ + 3OH-=Al(OH)3

1..3..78

x.y.m

1/x=78/m,3/y=78/m

x=m/78,y=3m/78

Al(OH)3+ OH-=AlO2 - + 2H2O

.1..27

.z.15m/26

1/z=27/15m/26

z=15m/702

n(AlCl)3=m/78

n(NaOH)=3m/78+15m/702

体积混合.所以浓度比=物质的量之比=m/78：（3m/78+15m/702）=9/42