1, Hoan thanh cac pt sau:
\(a.\)\(C4H9OH + 6O2 -> 4CO2 + 5H2O\)

\(b.\)\(COO H2n-2 +(\dfrac{n-1}{2})O2 -->CO2 + (n-1)H2O \)

\(c.\)\(2KMnO4 + 16HCl --> 2KCl +2MnCl2 + 5Cl2 + 8H2O \)

\(d.\) \( 2Al+ 3H_2SO_4 --> Al_2( SO_4)_3 +3H_2 \)

Bn có thể lm theo một trong hai cách 1 : 4CxHy + ( 4x + y) O2 --------> 4x CO2 + 2y H2O

Cách 2 : CxHy + (x + y/4) O2 -------> x CO2 + y/2 H2O

\(C_xH_y + (x + \dfrac{y}{4})O_2 \xrightarrow{t^o} xCO_2 + \dfrac{y}{2}H_2O\)

aj dup mjnh voi maj la ngay cuoi cung roi 

câu 1 :

a) 2Fe + 3Cl2  \(\rightarrow\)2FeCl3

b)2NaOH + H2SO4 \(\rightarrow\)NA2SO4 + 2H2O

c)Al2O3 + 6HCl \(\rightarrow\)2AlCl3 + 3H2O

d) SO2 + \(\frac{1}{2}\)O2 \(\rightarrow\)SO3

CÂU 2;

2I₂ + 5O2 ---->2I2O5

        0.875

nO2=0,875mol  ----->m I2O5=0.175*127=22.225 gam

V o2=0.875*22.4=19.6 lít

Câu 1: 

a) \(C_4H_9OH+6O_2\xrightarrow[]{t^o}4CO_2+5H_2O\)

b) \(2C_nH_{2n-2}+\left(3n-1\right)O_2\xrightarrow[]{t^o}2nCO_2+\left(2n-2\right)H_2O\)

c) \(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

Câu 2:

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\end{matrix}\right.\)

Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\\n_{MgO}=n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{rắn}=m_{Fe_2O_3}+m_{MgO}=0,1\cdot160+0,1\cdot40=20\left(g\right)\)

Câu 3: 

a) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)  (1)

                \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)  (2)

                \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)  (3)

b) Ta có: \(n_{H_2\left(3\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\) \(\Rightarrow m_{Cu}=12-0,1\cdot56=6,4\left(g\right)\) \(\Rightarrow n_{Cu}=0,1\left(mol\right)\)

Theo các PTHH: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=0,1\left(mol\right)\\n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,1\cdot80=8\left(g\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{CuO}=\dfrac{8}{8+8}\cdot100\%=50\%=\%m_{Fe_2O_3}\)

c) Theo các PTHH: \(n_{H_2}=n_{CuO}+3n_{Fe_2O_3}=0,1+0,15=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\) 

$C_nH_{2n} + \dfrac{3n}{2}O_2 \xrightarrow{t^o} nCO_2 + nH_2O$
$C_nH_{2n+3}N + \dfrac{3n+1,5}{2}O_2 \xrightarrow{t^o} nCO_2 + (n + 1,5)H_2O + 0,5N_2$

$C_nH_{2n+2}O + \dfrac{3n}{2}O_2 \xrightarrow{t^o} nCO_2 + (n + 1)H_2O$

$Fe + CuSO_4 \to FeSO_4 + Cu$
$4Al(NO_3)_3 \xrightarrow{t^o} 2Al_2O_3 + 12NO_2 + 3O_2$
$3Fe_3O_4 + 8Al \xrightarrow{t^o} 9Fe + 4Al_2O_3$

$3Fe_xO_y + 2yAl \xrightarrow{t^o} yAl_2O_3 + 3xFe$

1. \(C_nH_{2n}+\frac{3n}{2}O_2\rightarrow nCO_2+nH_2O\)

2.\(C_nH_{2n}+3N+\frac{3n}{2}O_2\rightarrow nCO_2+nH_2O+\frac{3}{2}N_2\)

3.\(C_nH_{2n}+2O+\frac{3n-2}{2}O_2\rightarrow nCO_2+nH_2O\)

4.\(Fe+CuSO_4\rightarrow FeSO_4+H_2O\)(Phương trình lỗi rồi)

5.\(Al\left(NO_3\right)_3\rightarrow Fe_2O_3+NO_2+O_2\)(Lỗi tiếp)

6.\(3Fe_3O_4+8Al\rightarrow4Al_2O_3+9Fe\)

7.\(3Fe_xO_y+2yAl\rightarrow3xFe+yAl_2O_3\)

Chọn đáp án C.