tham khảo

a) Gọi \(n_{N_2\left(pư\right)}=a\left(l\right)\left(a\le5\right)\)

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

            a---->3a---------->2a

`=>` \(B:\left\{{}\begin{matrix}N_2:5-a\left(l\right)\\H_2:10-3a\left(l\right)\\NH_3:2a\left(l\right)\end{matrix}\right.\)

`=>` \(5-a+10-3a+2a=10\)

`=> a = 2,5 (l)`

`=>` \(\left\{{}\begin{matrix}\%V_{N_2\left(dư\right)}=\dfrac{5-2,5}{10}.100\%=25\%\\\%V_{H_2\left(dư\right)}=\dfrac{10-3.2,5}{10}.100=25\%\\\%V_{NH_3}=100\%-25\%-25\%=50\%\end{matrix}\right.\)

b) Xét tỉ lệ: \(\dfrac{V_{N_2}}{1}>\dfrac{V_{H_2}}{3}\left(\dfrac{5}{1}>\dfrac{10}{3}\right)\Rightarrow\) Hiệu suất phản ứng tính theo `H_2`

`=>` \(H=\dfrac{3.2,5}{10}.100\%=75\%\)

c) Giá sử có 1 mol B, do các khí đo ở cùng đk 

`=>` \(\left\{{}\begin{matrix}n_{N_2}=0,25\left(mol\right)\\n_{H_2}=0,25\left(mol\right)\\n_{NH_3}=0,5\left(mol\right)\end{matrix}\right.\)

`=>` \(M_B=\dfrac{0,25.28+0,25.2+0,5.17}{1}=16\left(g/mol\right)\)

`=>` \(d_{B/kk}=\dfrac{16}{29}\approx0,552\)

cảm ơn ạ

1. \(2P+5H_2SO_4\rightarrow P_2O_5+5SO_2\uparrow+5H_2O\)

2. \(C+4HNO_3\rightarrow CO_2\uparrow+4NO_2\uparrow+2H_2O\)

3. \(S+6HNO_3\rightarrow H_2SO_4+6NO_2\uparrow+2H_2O\)

4. \(P+5HNO_3\rightarrow H_3PO_4+5NO_2\uparrow+H_2O\)

5. \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2\uparrow+2H_2O\)

6. \(C+2H_2SO_4\rightarrow CO_2\uparrow+2SO_2\uparrow+2H_2O\)

7. \(2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

8. \(2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2\uparrow+4H_2O\)

9. \(2Fe_3O_4+10H_2SO_4\rightarrow3Fe_2\left(SO_4\right)_3+SO_2+10H_2O\)

10. \(Fe+6HNO_3\rightarrow Fe\left(NO_3\right)_3+3NO_2\uparrow+3H_2O\)

11. \(8Fe+30HNO_3\rightarrow8Fe\left(NO_3\right)_3+3N_2O\uparrow+15H_2O\)

12. \(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO\uparrow+2H_2O\)

13. \(8Fe+30HNO_3\rightarrow8Fe\left(NO_3\right)_3+3NH_4NO_3+9H_2O\)

14. \(Fe_3O_4+10HNO_3\rightarrow3Fe\left(NO_3\right)_3+NO_2\uparrow+5H_2O\)

15. \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2\uparrow+8H_2O\)

16. \(3Fe_3O_4+28HNO_3\rightarrow9Fe\left(NO_3\right)_3+NO\uparrow+14H_2O\)

17. \(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO\uparrow+4H_2O\)

18. \(2Fe_xO_y+\left(6x-2y\right)H_2SO_4\rightarrow xFe_2\left(SO_4\right)_3+\left(3x-2y\right)SO_2\uparrow+\left(6x-2y\right)H_2O\)

19. \(K_2Cr_2O_7+14HCl\rightarrow2KCl+2CrCl_3+3Cl_2\uparrow+7H_2O\)

20. \(8A+10nHNO_3\rightarrow8A\left(NO_3\right)_n+nN_2O\uparrow+5nH_2O\)

21. \(10Fe+36HNO_3\rightarrow10Fe\left(NO_3\right)_3+3N_2\uparrow+18H_2O\)

22. \(Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2\uparrow+2H_2O\)

23. \(4Mg+10HNO_3\rightarrow4Mg\left(NO_3\right)_2+NH_4NO_3+3H_2O\)

24. \(3Zn+8HNO_3\rightarrow3Zn\left(NO_3\right)_2+2NO\uparrow+4H_2O\)

25. \(8Al+15H_2SO_4\rightarrow4Al_2\left(SO_4\right)_3+3H_2S\uparrow+15H_2O\)

tach ra di nhin lag mat lam

axit sulfuric

axit axetic

H2SO4 LÀ: axit sunfuric

CH3COOH LÀ : axit axetic

Gọi \(n_{CuSO_4}=n_{CaCO_3}=a\left(mol\right)\)

`=> 160a + 100a = 52`

`=> a = 0,2 (mol) = n_{CuSO_4} + n_{CaCO_3}`

khi cân bằng số nguyên tử mỗi nguyên tố ta viết......hệ số ........trước các công thức hóa học phải.......là bội số chung nhỏ nhất của.........các công thức hh.

a, \(2xFe+yO_2\underrightarrow{t^o}2Fe_xO_y\)

b, \(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)

a, 

b, 

mình mới lớp  7  nên sai sót gì mong bỏ qua

a) Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{HCl}=\dfrac{400.7,3\%}{36,5}=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\Rightarrow HCl\) dư, `Al` hết

b) Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

`=>` \(m_{ctan}=\left(0,8-0,6\right).36,5+0,2.133,5=34\left(g\right)\)

c) Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

`=> V_{H_2} = 0,3.22,4 = 6,72 (l)`

d) `m_{dd} = 5,4 + 400 - 0,3.2 = 404,6 (g)`

`=>` \(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{404,6}.100\%=6,6\%\\C\%_{HCl}=\dfrac{0,2.36,5}{404,6}.100\%=1,8\%\end{matrix}\right.\)

a)

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CaO + 2HCl \to CaCl_2 + H_2O$

Theo PTHH : 

$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$

b) $m_X = 10 : 44\% = \dfrac{250}{11}(gam)$
$m_{CaO} = \dfrac{250}{11}.56\% = \dfrac{140}{11}(gam)$
$n_{CaO} = \dfrac{140}{11} : 56 = \dfrac{5}{22}(mol)$
$n_{HCl} = 2n_{CaCO_3} + 2n_{CaO} = 0,655(mol)$

Câu 2 : 

Gọi $n_{CO_2} = a(mol) ; n_{H_2} = b(mol)$

Ta có : $a + b = \dfrac{3,36}{22,4} = 0,15$ và $\dfrac{44a + 2b}{a + b} = 8.2$

Suy ra : a = 0,05 ; b = 0,1

$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$Mg + 2HCl \to MgCl_2  + H_2$
Theo PTHH : 

$n_{MgCO_3} = n_{CO_2} = 0,05(mol) ; n_{Mg} = n_{H_2} = 0,1(mol)$
$m_{MgCO_3} = 0,05.84 = 4,2(gam) ; m_{Mg} = 0,1.24 = 2,4(gam)$