a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) - Pư hóa hợp

b, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) - Pư thế

c, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) - Pư thế

d, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) - Pư phân hủy.

a)     3Fe   +   2O2       Fe3O4   phản ứng hóa hợp 

b)     CuO   +   H2     Cu   +    H2O    phản ứng thế         

c)     2Al    +    6HCl       2AlCl3    +    3H2  phản ứng thế 

d)       2KMnO4       K2MnO4   +    MnO2   +    O2  phản ứng phân hủy

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Ta có: \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

a, Theo PT: \(n_{FeSO_4}=n_{Fe}=0,05\left(mol\right)\Rightarrow m_{FeSO_4}=0,05.152=7,6\left(g\right)\)

b, Theo PT: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{Cu\left(LT\right)}=n_{H_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{Cu\left(TT\right)}=0,05.64=3,2\left(g\right)\)

Mà: m_{Cu (TT)} = 3,04 (g)

\(\Rightarrow H\%=\dfrac{3,04}{3,2}.100\%=95\%\)

PT: $Fe+H^{2}S{O}^4\to FeS{O}^4+H^2$

Ta có: $n^{Fe}=\genfrac{}{}{0ex}{}{2,8}{56}=0,05\left(mol\right)$

a, Theo PT: $n^{FeS{O}^4}=n^{Fe}=0,05\left(mol\right)\Rightarrow m^{FeS{O}^4}=0,05.152=7,6\left(g\right)$

b, Theo PT: $n^{H^2}=n^{Fe}=0,05\left(mol\right)\Rightarrow V^{H^2}=0,05.22,4=1,12\left(l\right)$

c, PT: $CuO+H^2\underrightarrow{t^o}Cu+H^{2}O$

Theo PT: $n^{Cu\left(LT\right)}=n^{H^2}=0,05\left(mol\right)$

$\Rightarrow m^{Cu\left(TT\right)}=0,05.64=3,2\left(g\right)$

Mà: mCu (TT) = 3,04 (g)

$\Rightarrow H\%=\genfrac{}{}{0ex}{}{3,04}{3,2}.100\%=95\%$

a) $2Mg+O^2\stackrel{t^o}{\to }MgO$ - pư hóa hợp

b) $F{e}^3{O}^4+4H^2\stackrel{t^o}{\to }3Fe+4H^{2}O$ - pư thế

c) $Fe+H^{2}S{O}^{4\left(l\right)}\to FeS{O}^4+H^2$ - pư thế

d) $2KCl{O}^3\stackrel{t^o}{\to }2KCl+3O^2$ - pư phân hủy

Ta có: \(\dfrac{m_M}{m_N}=\dfrac{2}{5}\Rightarrow5m_M-2m_N=0\left(1\right)\)

Mà: m_M + m_N = 1 (tấn) = 1000 (kg) (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_M=\dfrac{2000}{7}\left(kg\right)\\m_N=\dfrac{5000}{7}\left(kg\right)\end{matrix}\right.\)

\(\Rightarrow m_{Fe_2O_3}=m_M.60\%=\dfrac{1200}{7}\left(kg\right)\Rightarrow n_{Fe_2O_3}=\dfrac{\dfrac{1200}{7}}{160}=\dfrac{15}{14}\left(kmol\right)\)

\(\Rightarrow n_{Fe}=2n_{Fe_2O_3}=\dfrac{15}{7}\left(kmol\right)\)

\(m_{Fe_3O_4}=m_N.69,6\%=\dfrac{3480}{7}\left(kg\right)\Rightarrow n_{Fe_3O_4}=\dfrac{\dfrac{3480}{7}}{232}=\dfrac{15}{7}\left(kmol\right)\)

\(\Rightarrow n_{Fe}=3n_{Fe_3O_4}=\dfrac{45}{7}\left(kmol\right)\)

\(\Rightarrow m_{Fe}=\left(\dfrac{15}{7}+\dfrac{45}{7}\right).56=480\left(kg\right)\)

Bạn xem lời giải ở đây nhé.

https://olm.vn/hoi-dap/detail/7712766586263.html

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Có: O₂ hao hụt 40% → H% = 100 - 40 = 60%

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,4}{60\%}=\dfrac{2}{3}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=\dfrac{2}{3}.158\approx105,3\left(g\right)\)

\(M_A=5.2=10\left(g/mol\right)\)

Do các khí đo ở cùng điều kiện nhiệt độ và áp suất nên tỉ lệ thể tích cũng là tỉ lệ mol

Chọn \(\left\{{}\begin{matrix}n_{H_2}=17,5\left(mol\right)\\n_{N_2}=5\left(mol\right)\end{matrix}\right.\)

Gọi \(n_{N_2\left(p\text{ư}\right)}=a\left(mol\right)\left(0< a< 5\right)\)

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

             a---->3a---------->2a

Xét tỉ lệ: \(5< \dfrac{17,5}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂

Ta có: \(n_A=5+17,5+2a-a-3a=22,5-2a\left(mol\right)\)

Theo ĐLBTKL: \(m_A=5.28+17,5.2=175\left(g\right)\)

\(\Rightarrow M_A=\dfrac{175}{22,5-2a}=10\Leftrightarrow a=2,5\left(TM\right)\)

\(\Rightarrow H=\dfrac{2,5}{5}.100\%=50\%\)

Kali clorat chứ bạn:v?

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: \(2KClO_3\xrightarrow[]{MnO_2,t^o}2KCl+3O_2\)

              \(\dfrac{1}{3}\)<---------------------------0,5

\(\Rightarrow m=m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)

giúp mk vs các bn 

Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=2a\left(mol\right)\\n_{Fe}=3a\left(mol\right)\end{matrix}\right.\)

Ta có: \(m_{t\text{ăng}}=m_{KL}-m_{H_2}\)

\(\Rightarrow m-m_{H_2}=m-2,4\\ \Leftrightarrow m_{H_2}=2,4\left(g\right)\Rightarrow n_{H_2}=\dfrac{2,4}{2}=1,2\left(mol\right)\)

PTHH:
`Mg + 2HCl -> MgCl_2 + H_2`

`Zn + 2HCl -> ZnCl_2 + H_2`

`Fe + 2HCl -> FeCl_2 + H_2`

Theo PTHH: 

\(n_{H_2}=n_{Mg}+n_{Zn}+n_{Fe}=a+2a+3a=6a\left(mol\right)\\ \Rightarrow6a=1,2\Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Zn}=0,2.2=0,4\left(mol\right)\\n_{Fe}=0,2.3=0,6\left(mol\right)\end{matrix}\right.\)

Vậy \(m=0,2.24+0,4.65+0,6.56=64,4\left(g\right)\)

$$\begin{array}{c}{\Delta }_m=m_Y-m_{H_2}\\ \Rightarrow m-2,4=m-m_{H_2}\\ \iff m_{H_2}=2,4gam\\ \Rightarrow n_{H_2}=\frac{2,4}{2}=1,2mol\end{array}$$

Gọi n_Mg là A => n_Zn là 2a, n_Fe là 3a

$$\begin{array}{c}Mg+2HCl\to MgC{l}_2+H_2\\ \mathrm{a\; a}\end{array}$$

$$\begin{array}{c}Zn+2HCl\to ZnC{l}_2+H_2\\ \mathrm{2a\; 2a}\end{array}$$$$\begin{array}{c}Fe+2HCl\to FeC{l}_2+H_2\\ 3a\; 3a\end{array}$$

$$\begin{array}{c}\Rightarrow n_{H_2}=a+2a+3a=1,2mol\\ \Rightarrow a=0,2mol\\ \Rightarrow m=0,2\cdot 24+0,2\cdot 2\cdot 65+0,2\cdot 3\cdot 56=64,4gam\end{array}$$