Tham khảo
Ta có pthh
2Al + 3H2SO4 →→ Al2(SO4)3 + 3H2 (1)
Mg + H2SO4 →→ MgSO4 + H2 (2)
Theo đề bài ta có
VH2=1568 ml =1,568 l
-> nH2=mct.100%C%=6,68.100%1,96%=350gmct.100%C%=6,68.100%1,96%=350g
mdd(sau-phan-ung) = mhh + mddH2SO4 - mH2 = 1,41 +350 - (0,07.2)=351,27 g
Theo pthh 1
nAl2(SO4)3=1/2nAl=1/2.0,03=0,015 mol
->mAl2(SO4)3=0,015.342=5,13 g
Theo pthh 2
nMgSO4=nMg=0,025 mol
->mMgSO4=0,025.120=3g
⇒⇒ C%3351,27.100%≈0,854%