Bài tập Hóa Học - Bão toàn mol electron

By Trang Ánh Nam -

hoà tan hoàn toàn 9g hỗn hợp x gồm mg ,al vào dung dịch h2so4 (loãng dư) thu được 10,08l khí h2 .mặt khác, cho 4,5g x vào dung dịch hno3 thu được 1,12l khí No duy nhất (đktc) và dung dịch chứa m gam hỗn hợp muối ,a : viết pthh , b:m=?
giải:
Mg-->Mg+2+2e
Al --->Al+3 +3e
Khi phản ứng với H2SO4
2H +2e -->H2.
…….0,9……0,45
Khi phản ứng với HNO3
N+5+3e -->N+2.
……0,15….0,05
Vì số mol e lúc này phải là 0,45 vì lấy một nữa lượng chất ở trên, nên ta còn 0,3 mol e tạo thành NH4NO3.
N+5 +8e -->N-3.
…….0,3……0,0375
Vậy :
m=mMg(NO3)2+mAl(NO3)3+mNH4NO3
=mkl+mNO3++mNH4NO3
=4,5+0,45.62+0,0375.80
=4,5+27,9+3
=35,4 g.

ở đây ta lưu ý số mol NO3 trong muối Mg(NO3)2 và Al(NO3)3 bằng sô mol mà kim loại trao đổi. (phần T tô đỏ)

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