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Question

cho 5,2g lim loai R vao dung dich HNO3 15.75% vua du de hoa tan het kim loai duoc dung dich X va 1.008 lit hon hop 2 khi NO va N2O (dktc) sau phan ung khoi luong dung dich tang 3,78g so voi ban dau
a/ xac dinh kim loai R
b/ tinh C% muoi trong X

viet130480 · Answer 1 · 11/12/2012 2:30:10 AM

Bình chọn tăng 1 Bình chọn giảm

a)Ta có:
$\begin{cases}n_{NO}+n_{N_2O}=0,045 \\ 30n_{NO}+44n_{N_2O} =5,2-3,78 \end{cases} \Rightarrow \begin{cases}n_{NO}=0,04 mol \\ n_{N_2O}=0,005 mol \end{cases}$
Lại có:
$M-ne\rightarrow M^{n+}$
$N^{+5}+3e\rightarrow NO$
$2N^{+5}+8e\rightarrow N_2O$
Áp dụng định luật bảo toàn electron ta có :$n.n_M=0,04.3+0,005.8=0,16 mol$
$\Rightarrow n.\frac{5,2}{M}=0,16\Rightarrow M=32,5n\Rightarrow $ lập bảng $\Rightarrow n=2\Rightarrow M=65\Rightarrow M:Zn$
b) $n_{HNO_3}=2n_{Zn}+n_{NO}+2n_{N_2O}=0,21 mol$
$\Rightarrow m_{dd}=\frac{0,21.63.100}{15,75}+3,78=87,78 gam\Rightarrow C\%=\frac{0,08.189.100}{87,78}=17,22\%$

Trả lời 12-11-12 02:30 AM