Công thức tính nhanh số mol H2 sinh ra của 1 chất:

\({\text{n}}_{H_2^{}}^{} = \frac{{n(HoaTri)}}{2}.n_{KL}^{}\)

\(Mg:x;Fe:y\)

\(24x + 56y = 22 - 7,2\)

\(x + y = 0,35\)

\( \to x = 0,15;y = 0,2\)

\(\% m_{Mg}^{} = \frac{{0,15.24}}{{22}}.100 = 16,36\% \)

\(\% m_{Fe}^{} = \frac{{0,2.56}}{{22}}.100 = 50,9\% \)

\(\% m_{Ag}^{} = \frac{{7,2}}{{22}}.100 = 32,74\% \)

\(b)Mg \to M{g^{2 + }} + 2e\)

\({\text{   0}}{\text{,15                0}}{\text{,3}}\)

\(Fe \to F{e^{2 + }} + 2e\)

\({\text{ 0}}{\text{,2              0}}{\text{,4}}\)

\(Ag \to A{g^ + } + 1e\)

\({\text{ }}\frac{1}{{15}}{\text{              }}\frac{1}{{15}}\)

\({N^{ + 5}} + 1e \to N_{}^{ + 4}\)

\({\text{x         x}}\)

\({\text{Bao toan electron: n}}_{e_{Cho}^{}}^{} = n_{e_{Nhan}^{}}^{}\)

\(x = \frac{1}{{15}} + 0,4 + 0,3 = \frac{{23}}{{30}}\;{\text{mol}}\)

\(V_{NO_2^{}}^{} = \frac{{23}}{{30}}.22,4 = 17,17l\)

\(2.\)

\(n_{H_2^{}}^{} = 0,75\;{\text{mol}}\)

\(TN1:n_{Al}^{} = \frac{2}{3}.n_{H_2^{}}^{} = 0,5{\text{ mol}}\)

\({\text{TN2:}}\)

\(Al \to A{l^ 3+ } + 3e\)

\({\text{ 0}}{\text{,5             1}}{\text{,5}}\)

\(Cu \to Cu_{}^{2 + } + 2e\)

\({\text{ x                  2x}}\)

\({N^{ + 5}} + 1e \to N_{}^{ + 4}\)

\({\text{          2       2}}\)

\({\text{BTE: 1}}{\text{,5  +  2x  =  2}}\)

\( \to {\text{x  =  0}}{\text{,25}}\)

\(m = 0,25.64 + 0,5.27 = 29,5g\)

Đáp án:

 Bạn tham khảo nhé 

Giải thích các bước giải:

 Bài 1: 

`a)`

`hh: n_{Mg} = x(mol); n_{Fe} = y(mol); n_{Ag} = 7,2(g)`

`Fe → Fe^{+2} + 2e`

`Mg → Mg^{+2} + 2e`

`2H^{+} + 2e → H_2`

`n_{H_2} = {7,84}/{22,4} = 0,35(mol)`

`m = 56y + 24x = 22 - 7,2(g)(1)`

Bảo toàn `e`: `2x + 2y = 0,35. 2(2)`

`(1)(2) → x = 0,15(mol); y = 0,2(mol)`

`%m_{Ag} = {7,2}/{22} .100 = 32,73%`

`%m_{Fe} = {0,2.56}/{22} .100 = 50,9%`

`%m_{Mg} = 100-50,9-32,73= 16,37%`

`b)`

`Fe → Fe^{+3} + 3e`

`Ag → Ag^{+} + 1e`

`Mg → Mg^{+2} + 2e`

`N^{+5} + 1e → N^{+4}`

`n_{Ag} = {7,2}/{108} = 1/{15} (mol)`

Bảo toàn `e`: 

`n_{NO_2} = 1/{15} + 3.  0,2 + 2.  0,15= {29}/{30} (mol)`

`V_{NO_2} = {29}/{30}    . 22,4 = 21,653(l)`

Bài 2: 

`n_{Cu} = x(mol); n_{Al} = y(mol)`

`H_2SO_4`: 

`Al → Al^{+3} + 3e`

`2H^{-} + 2e → H_2`

`n_{H_2} = {16,8}/{22,4} = 0,75(mol)`

Bảo toàn `e: 3y = 0,75.2(1)`

`HNO_3`: 

`Cu → Cu^{+2} + 2e`

`Al → Al^{+3} + 3e`

`N^{+5} + 1e → N^{+4}`

`n_{NO_2} = {44,8}/{22,4} = 2(mol)`

Bảo toàn `e`: `2x + 3y = 2(2)`

`(1)(2) → x = 0,25(mol); y = 0,5(mol)`

`m = 0,25 .64 + 0,5 .27 = 29,5(g)`