giúp tôi với 

mai nộp cô 

1 . 

\(nFe=11,2\div56=0,2\left(mol\right)\)

\(nAi=\frac{m}{27}\)\(\left(mol\right)\)

Khi them Fe vao coc dung dd HCl ( coc A ) co phan ung :

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

Theo dinh luat bao toan nang luong , khoi luong coc dung HCl tang them :

\(2Al+3H_2SO_4\Rightarrow Al_2\left(SO_4\right)_3+3H_2\)

- Khi cho m gam vao Al vao coc B , coc B tang them m \(\frac{3.m}{27.2}\)

- De can thang bang khoi luong o coc dung H2SO4 , cung phai tang them 10,8 g  . co 

\(m=12,15\left(g\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

\(n_{H_2}=\frac{0,896}{22,4}=0,04mol\)

\(\hept{\begin{cases}27a+56b=1,1\\1,5a+b=0,04\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=0,02\\b=0,01\end{cases}}\)

\(m_{Al}=0,02.27=0,54g\)

\(m_{Fe}=1,1-0,54=0,56g\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

\(n_{H_2}=\frac{0,896}{22,4}=0,04mol\)

\(\hept{\begin{cases}27a+56b=1,1\\1,5a+b=0,04\end{cases}}\)

\(\rightarrow\hept{\begin{cases}a=0,02\\b=0,01\end{cases}}\)

\(m_{Al}=0,02.27=0,54g\)

\(m_{Fe}=1,1-0,54=0,56g\)

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

\(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)

Theo phương trình \(n_{Al}=n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,2mol\)

\(\rightarrow m_{Al}=0,2.27=5,4g\)

\(\rightarrow m=5,4\)

b. \(m_{\text{muối}}=m_{AlCl_3}=0,2.133,5=26,7g\)

a)PTHH\(2AL+6HCL\rightarrow2ALCL_3+3H_2\uparrow\)

\(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)

Theo phương trình:\(n_{AL}=n_{alcl_3}=\frac{2}{3}n_{H_2}=0,2mol\)

\(\rightarrow m_{AL}=0,2\cdot27=5,4g\)

\(\rightarrow m=5,4\)

b)\(m_{muối}=m_{alcl_3}=0,2\cdot133,5=26,7g\)

what dịch là cái gì
 

what là cái gì

a. \(ZnCl_2+Zn^{2+}+2Cl^-\)

b. \(FeSO_4\rightarrow Fe^{2+}+SO_4^{2-}\)

c. \(Zn\left(NO_3\right)_2\rightarrow Zn^{2+}+2NO_3^-\)

d. \(MgCl_2\rightarrow Mg^{2+}+2Cl^-\)

a)PTHH:\(2KCLO_3\rightarrow^{t^o}2KCL+3O_2\)

b)\(n_{KCL}=\frac{2,98}{74,5}=0,4mol\)

Từ phương trình hóa học\(n_{KCL}=n_{KCLO_3}=0,04mol\)

\(n_{KCLO_3}=0,04\cdot122,5=4,9g\)

\(H=\frac{4,9}{9,8}\cdot100\%=50\%\)

a. PTHH: \(2KClO_3\rightarrow^{t^o}2KCl+3O_2\)

b. \(n_{KCl}=\frac{2,98}{74,5}=0,4mol\)

Từ phương trình hoá học \(n_{KCl}=n_{KClO_3}=0,04mol\)

\(n_{KClO_3}=0,04.122,5=4,9g\)

\(H=\frac{4,9}{9,8}.100\%=50\%\)

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