Bài 7: Nitơ

\(Cu\rightarrow Cu^{2+}+2e\\ 4H^++NO_3^-+3e\rightarrow NO+20H_2O\\ CuO+HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\\ BTe:n_{NO}.3=n_{Cu}.2\\ \Rightarrow n_{Cu}=0,3\left(mol\right)\\ \Rightarrow n_{H^+}=n_{HNO_3\left(pứCu\right)}=0,2.4=0,8\left(mol\right)\\ \Rightarrow n_{HNO_3\left(pứCuO\right)}=1.1-0,8=0,2\left(mol\right)\\ Tacó:n_{CuO}=n_{HNO_3\left(pứCuO\right)}=0,2\left(mol\right)\\ \Rightarrow m=m_{Cu}+m_{CuO}=0,3.64+0,2.80=35,2\left(g\right)\)

Bảo toàn nguyên tố N: \(n_{NH_3}=n_{HNO_3}=1\left(mol\right)\\ \Rightarrow m_{NH_3}=1.17=17\left(kg\right)\)

=> Chọn B

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

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

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

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 

`=>` \(H=\dfrac{0,5}{2}.100\%=25\%\)

`(1) 4NO_2 + 2H_2O + O_2 -> 4HNO_3`

`(2) HNO_3 + CuO -> Cu(NO_3)_2 + H_2O`

`(3) Cu(NO_3)_2 + 2KOH -> Cu(OH)_2 + 2KNO_3`

`(4) Cu(OH)_2 + 2HNO_3 -> Cu(NO_3)_2 + 2H_2O`

$(5) 2Cu(NO_3)_2 \xrightarrow{t^o} 2CuO + 4NO_2 + O_2$

$(6) CuO + CO \xrightarrow{t^o} Cu + CO_2$

N2O

NO2

NO

N2O5

NH3

NH4NO3

Gọi \(\left\{{}\begin{matrix}n_{N_2}=x\left(mol\right)\\n_{H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{28x+2y}{x+y}=12,4\)

\(\Leftrightarrow12,4x+12,4y-28x-2y=0\Leftrightarrow-15,6x=-10,4y\\ \Leftrightarrow\dfrac{x}{y}=-\dfrac{10,4}{-15,6}=\dfrac{2}{3}\)

Giả sử \(\left\{{}\begin{matrix}n_{N_2}=2\left(mol\right)\\n_{H_2}=3\left(mol\right)\end{matrix}\right.\)

\(n_{khí.giảm}=n_{NH_3}=\dfrac{2}{3}n_{H_2}.H\%=\dfrac{2}{3}.3.40\%=0,8\left(mol\right)\)

\(n_Y=2+3-0,8=4,2\left(mol\right)\)

\(m_Y=m_X=28.2+2.3=62\left(g\right)\)

\(\overline{M_Y}=\dfrac{62}{4,2}=14,76g/mol\)

\(n_{N_2}=a;n_{H_2}=b\\ PT:\dfrac{1}{2}N_2+\dfrac{3}{2}H_2-Fe,t^{^0}->NH_3\\ M_X=\dfrac{28a+2b}{a+b}=12,4\\ a=1,5b\\ Sau.pư:\\ M_Y=\dfrac{2.0,6b+28\left(a-\dfrac{0,4b}{3}\right)+17.0,4b}{0,6b+a-\dfrac{0,4b}{3}+0,4b}=19,55\)

\(n_{N_2}=a,n_{H_2}=b\\ M_A=\dfrac{28a+2b}{a+b}=7,2\\ a=4b\\ PT:\dfrac{1}{2}N_2+\dfrac{3}{2}H_2-Fe,t^{^0}->NH_3\\ CuO+H_2-t^{^0}->Cu+H_2O\\ n_{Cu}=n_{H_2dư}=\dfrac{32,64}{64}=0,51mol\\ n_{H_2pư}=b-0,51\left(mol\right)\\ H=\dfrac{b-0,51}{b}=0,2\\ b=0,6375\\ a=2,55\\ V_A=22,4\left(0,6375+2,55\right)=71,4L\)

Từ pV = nRT, thấy rằng khi p giảm 5% thì n giảm 5%.

\(n_{N_2}=a;n_{H_2}=b;n_{N_2pư}=0,1a\\ PT:\dfrac{1}{2}N_2+\dfrac{3}{2}H_2-Fe,t^{^0}->NH_3\\ n_{sau}=0,95\left(a+b\right)=0,1a+0,3a+0,2a=0,6a\\ a=2,714b\\ \%V_{N_2}=\dfrac{a}{a+b}.100\%=73,07\%\\ \%V_{H_2}=26,93\%\)