一道化学题,求详解~!!感激

2024-12-24 14:05:27
推荐回答(3个)
回答1:

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

8.96L=0.25mol

列比例解得:NA=4.6g,Al=5.3g

c=0.2mol/0.2L=1mol/L

回答2:

23n(Na)+27n(Al)=10
n(Na)+3n(Al)=8.96/11.2
4.6 5.4 1mol/L

回答3:

2Na + 2H2O = 2NaOH + H2↑
2Al + 2NaOH + 2H2O = 2NaAlO2 + 3H2↑
(1)设钠和铝各为xmol,ymol
所以23X+27Y=10
1/2X+3/2Y=8.96/22.4
X=0.2mol Y=0.2mol
mNa=4.6g mAl=5.4g
(2)因为恰好反映所以溶质为n NaAlO2=nNa=0.2mol
c=0.2mol/0.2L=1mol/L