Câu 10:

Ta có: \(n_{NH_3}=\dfrac{6,8}{17}=0,4\left(mol\right)\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

             1------------------->2

`=>` \(H=\dfrac{0,4}{2}.100\%=20\%\)

Câu 11:

Ta có: \(\left\{{}\begin{matrix}n_{N_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\n_{H_2}=\dfrac{39,2}{22,4}=1,75\left(mol\right)\\n_{NH_3}=\dfrac{3,4}{17}=0,2\left(mol\right)\end{matrix}\right.\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

            0,1<-0,3<---------0,2

Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{1,75}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂ 

`=>` \(H=\dfrac{0,1}{0,5}.100\%=20\%\)

Câu 12:

Gọi \(n_{N_2}=a\left(l\right)\left(ĐK:0< a< 2\right)\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

            a----->3a--------->2a

Sau phản ứng có: \(\left\{{}\begin{matrix}n_{N_2}=2-a\left(l\right)\\n_{H_2}=8-3a\left(l\right)\\n_{NH_3}=2a\left(l\right)\end{matrix}\right.\)

`=> 2 - a + 8 - 3a + 2a = 9`

`=> a = 0,5 (l) (TM)`

`=>` \(V_{NH_3}=2.0,5=1\left(l\right)\)

b) Xét tỉ lệ: \(\dfrac{2}{1}< \dfrac{8}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂ 

`=>` \(H=\dfrac{0,5}{2}.100\%=25\%\)

Câu 2:

a)

(1) \(NH_4NO_2\xrightarrow[]{t^o}N_2\uparrow+2H_2O\)

(2) \(NH_4NO_2+NaOH\rightarrow NaNO_2+NH_3\uparrow+H_2O\)

(3) \(N_2+O_2\xrightarrow[]{3000^oC}2NO\)

(4) \(NH_4NO_2+Mg\xrightarrow[]{t^o}Mg_3N_2+2H_2O\)

(5) \(2NO+O_2\rightarrow2NO_2\)

(6) \(Mg_3N_2\xrightarrow[]{t^o}3Mg+N_2\)

(7) \(2NH_3+3Cl_2\rightarrow N_2+6HCl\)

b)

\(NH_4NO_2\xrightarrow[]{t^o}N_2\uparrow+2H_2O\)

$N_2 + 3H_2 \xrightarrow{t^o,p,xt} 2NH_3$

`2NH_3 + H_2SO_4 -> (NH_4)_2SO_4`

$(NH_4)_2SO_4 + BaCl_2 \rightarrow 2NH_4Cl + BaSO_4 \downarrow$
$NH_4Cl + AgNO_3 \rightarrow AgCl \downarrow + NH_4NO_3$
$NH_4NO_3 \xrightarrow{t^o} N_2O + 2H_2O$
c)

\(NH_4NO_2\xrightarrow[]{t^o}N_2\uparrow+2H_2O\)

$N_2 + 3H_2 \xrightarrow{t^o,p,xt} 2NH_3$

$4NH_3 + 5O_2 \xrightarrow[Pt]{t^o} 4NO + 6H_2O$
$2NO + O_2 \rightarrow 2NO_2$
$2NH_3 + 3CuO \xrightarrow{t^o} 3Cu + 3H_2O + N_2$

$Cu + 2H_2SO_{4(đặc)} \xrightarrow{t^o} CuSO_4 + SO_2 + 2H_2O$
$2SO_2 + O_2 \xrightarrow[V_2O_5]{t^o} 2SO_3$
`SO_3 + H_2O -> H_2SO_4`

`2NH_3 + H_2SO_4 -> 2NH_4HSO_4`

`NH_4HSO_4 + NH_3 -> (NH_4)_2SO_4`

`(NH_4)_2SO_4 + H_2SO_4 -> 2NH_4HSO_4`

d) 

`NH_4NO_3 + NaOH -> NaNO_3 + NH_3 + H_2O`

$3NH_3 + 3H_2O + AlCl_3 \rightarrow Al(OH)_3 \downarrow + 3NH_4Cl$
`Al(OH)_3 + NaOH -> NaAlO_2 + 2H_2O`

$CO_2 + NaAlO_2 + 2H_2O \rightarrow Al(OH)_3 \downarrow + NaHCO_3$
$2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O$
Câu 3:

a) \(NaNO_2+NH_4Cl\xrightarrow[]{t^o}NaCl+N_2\uparrow+2H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}n_{NaNO_2}=0,2.1,5=0,3\left(mol\right)\\n_{NH_4Cl}=0,2.2=0,4\left(mol\right)\end{matrix}\right.\)

PTHH:              \(NaNO_2+NH_4Cl\xrightarrow[]{t^o}NaCl+N_2\uparrow+2H_2O\)

ban đầu             0,3              0,4

phản ứng           0,3---------->0,3

sau phản ứng     0                 0,1           0,3        0,3

                          \(NH_4Cl\xrightarrow[]{t^o}NH_3+HCl\)

                             0,1-------->0,1---->0,1

`=>` \(V_{khí\left(N_2.NH_3,HCl\right)}=\left(0,1+0,1+0,3\right).22,4=11,2\left(l\right)\)

`=>` \(C_{M\left(NaCl\right)}=\dfrac{0,3}{0,2+0,2}=0,75M\)

Gọi kim loại hóa trị II là R

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH:

`Fe + 2HCl -> FeCl_2 + H_2`

`R + 2HCl -> RCl_2 + H_2`

Theo PT: \(n_R+n_{Fe}=n_{H_2}=0,05\left(mol\right)\)

`=>` \(\overline{M_{R,Fe}}=\dfrac{2}{0,05}=40\left(g/mol\right)\)

Mà \(40< 56=M_{Fe}\)

`=>` \(M_R< 40< 56\left(1\right)\)

- Mặt khác: \(\left\{{}\begin{matrix}n_R=\dfrac{4,8}{M_R}\left(mol\right)\\n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\end{matrix}\right.\)

PTHH: `R + 2HCl -> RCl_2 + H_2`

Theo PT: \(n_{HCl\left(pư\right)}=2n_R=\dfrac{9,6}{M_R}< n_{HCl\left(bđ\right)}\)

`=>` \(\dfrac{9,6}{M_R}< 0,5\Leftrightarrow M_R>19,2\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow19,2< M_R< 40\)

Mà R có hóa trị II `=> R` là Magie (Mg)

Gọi số hạt p, n, e trong XY₂ là p, n, e

                            trong X là p_X, n_X, e_X

                            trong Y là p_Y, n_Y, e_Y

Theo bài ra, ta có: \(\left\{{}\begin{matrix}p+e+n=114\\p=e\\\dfrac{p+e}{n}=2\end{matrix}\right.\Leftrightarrow p=e=n=38\)

Ta lại có: \(\left\{{}\begin{matrix}p=p_X+2p_Y=38\\p_X=e_X\\p_Y=e_Y\\\dfrac{p_X+e_X}{p_Y+e_X}.100\%=37,5\%\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}p_X=e_X=6\\p_Y=e_Y=16\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}X:Cacbon\left(C\right)\\Y:Oxi\left(O\right)\end{matrix}\right.\)

A là CO₂

Ta có :

ab = ba - 63

10a + b = 10b + a - 63

9a = 9b - 63

⇒a = 1 ; b = 8

Tổng a và b là:

1 + 8 = 9

Theo ĐLBTNT: \(\left\{{}\begin{matrix}n_S=n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_H=2.n_{H_2O}=2.\dfrac{3,6}{18}=0,4\left(mol\right)\end{matrix}\right.\)

`=>` \(n_{O\left(A\right)}=\dfrac{6,8-0,2.32-0,4}{16}=0\left(mol\right)\)

`=>` A không chứa O

`=>` \(n_H:n_S=0,4:0,2=2:1\)

`=>` A là `H_2S`

Tên: hiđro sunfua

Cách 1: \(M_E=\dfrac{18}{0,3}=60\left(g/mol\right)\)

BTNT C: \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=20,4\left(g\right)\)

`=>` \(n_H=2.n_{H_2O}=2.\dfrac{20,4-0,3.44}{18}=0,8\left(mol\right)\)

`=>` \(n_{O\left(E\right)}=\dfrac{6-0,8-0,3.12}{16}=0,1\left(mol\right)\)

Ta có: \(n_C:n_H:n_O=0,3:0,8:0,1=3:8:1\)

`=>` CTPT của E có dạng \(\left(C_3H_8O\right)_n\)

`=>` \(n=\dfrac{60}{60}=1\)

`=> E` là `C_3H_8O`

Cách 2: \(M_E=\dfrac{18}{0,3}=60\left(g/mol\right)\)

`=>` \(n_E=\dfrac{6}{60}=0,1\left(mol\right)\)

BTNT C: \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=20,4\left(g\right)\)

`=>` \(n_H=2.n_{H_2O}=2.\dfrac{20,4-0,3.44}{18}=0,8\left(mol\right)\)

`=>` \(n_{O\left(E\right)}=\dfrac{6-0,8-0,3.12}{16}=0,1\left(mol\right)\)

Trong 1 mol E có: \(\left\{{}\begin{matrix}n_C=\dfrac{0,3}{0,1}=3\left(mol\right)\\n_H=\dfrac{0,8}{0,1}=8\left(mol\right)\\n_O=\dfrac{0,1}{0,1}=1\left(mol\right)\end{matrix}\right.\)

`=>` CTPT của E là `C_3H_8O`

\(M_D=\dfrac{14,8}{0,2}=74\left(g/mol\right)\)

BTNT C: \(n_C=n_{CO_2}=n_{BaCO_3}=\dfrac{78,8}{197}=0,4\left(mol\right)\)

Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=26,6\left(g\right)\)

`=>` \(n_H=2.n_{H_2O}=2.\dfrac{26,6-0,4.44}{18}=1\left(mol\right)\)

`=>` \(n_{O\left(D\right)}=\dfrac{7,4-0,4.12-1}{16}=0,1\left(mol\right)\)

Cách 1: 

Ta có: \(n_C:n_H:n_O=0,4:1:0,1=4:10:1\)

`=>` CTPT của E có dạng \(\left(C_4H_{10}O\right)_n\)

`=>` \(n=\dfrac{74}{74}=1\)

`=> D` là \(C_4H_{10}O\)

Cách 2:

Ta có: \(n_D=\dfrac{7,4}{74}=0,1\left(mol\right)\)

`=>` 1 mol D có: \(\left\{{}\begin{matrix}n_C=\dfrac{0,4}{0,1}=4\left(mol\right)\\n_H=\dfrac{1}{0,1}=10\left(mol\right)\\n_O=\dfrac{0,1}{0,1}=1\left(mol\right)\end{matrix}\right.\)

`=> D` là \(C_4H_{10}O\)

Sửa đề 12,6 gam => 12,4 gam

\(M_B=1,875.32=60\left(g/mol\right)\)

BTNT C: \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)

Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=12,6\left(g\right)\)

`=>` \(n_H=2.n_{H_2O}=2.\dfrac{12,4-0,2.44}{18}=0,4\left(mol\right)\)

`=>` \(n_{O\left(B\right)}=\dfrac{6-0,4-0,2.12}{16}=0,2\left(mol\right)\)

Cách 1: 

Ta có: \(n_C:n_H:n_O=0,2:0,4:0,2=1:2:1\)

`=>` CTPT của B có dạng `(CH_2O)_n`

`=>` \(n=\dfrac{60}{30}=2\)

Vậy B là `C_2H_4O_2`

Cách 2: \(n_B=\dfrac{6}{60}=0,1\left(mol\right)\)

`=>` trong 1 mol B có \(\left\{{}\begin{matrix}n_C=\dfrac{0,2}{0,1}=2\left(mol\right)\\n_H=\dfrac{0,4}{0,1}=4\left(mol\right)\\n_O=\dfrac{0,2}{0,1}=2\left(mol\right)\end{matrix}\right.\)

Vậy B là `C_2H_4O_2`