Question

[Nhật]

Làm bài nào cx đc nha

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

Bài 4:

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) n_HCl = 0,5.1 = 0,5 (mol)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => Zn hết, HCl dư

c) 

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

            0,2---->0,4------>0,2

\(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,5-0,4}{0,5}=0,2M\\C_{M\left(ZnCl_2\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)

Bài 5:

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

b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{100.49\%}{98}=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{0,5}{1}\) => Fe hết, H₂SO₄ dư

c)

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

            0,4---->0,4------->0,4--->0,4

\(\left\{{}\begin{matrix}m_{H_2SO_4\left(dư\right)}=\left(0,5-0,4\right).98=9,8\left(g\right)\\m_{FeSO_4}=0,4.152=60,8\left(g\right)\end{matrix}\right.\)

m_{dd sau pư} = 22,4 + 100 - 0,4.2 = 121,6 (g)

\(\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{9,8}{121,6}.100\%=8,06\%\\C\%_{FeSO_4}=\dfrac{60,8}{121,6}.100\%=50\%\end{matrix}\right.\)

Bài 6: 

\(n_{HCl}=1.0,6=0,6\left(mol\right)\)

PTHH: \(A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\)

              0,1<----0,6

=> \(M_{A_2O_3}=\dfrac{10,2}{0,1}=102\left(g/mol\right)\Rightarrow M_A=27\left(g/mol\right)\)

=> A là Al

CTHH: Al₂O₃

Answer 2

[kaito kid]

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