lỗi

Ủa sao ko có đầu bài zậy!!!

Bài 22:

a)

\(\left\{{}\begin{matrix}n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{294.10\%}{98}=0,3\left(mol\right)\end{matrix}\right.\)

b)

PTHH: `MgO + H_2SO_4 -> MgSO_4 + H_2O`

bđ         0,2          0,3

pư         0,2----->0,2

spư        0          0,1              0,2

\(m_{MgSO_4}=0,2.120=24\left(g\right);m_{H_2SO_4.dư}=0,1.98=9,8\left(g\right)\)

c)

\(m_{dd.sau.pư}=8+294=302\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{24}{302}.100\%=7,95\%\\C\%_{H_2SO_4.dư}=\dfrac{9,8}{302}.100\%=3,25\%\end{matrix}\right.\)

B25:

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(m_{HCl}=\dfrac{109,5.10\%}{100}=10,95\left(g\right)\)\(\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,1       0,2         0,1          0,1  (mol)

LTL : \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) => Mg đủ, HCl dư

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=2,4+109,5-\left(0,1.2\right)=111,7\left(g\right)\)

\(m_{HCl.dư.sau.pứ}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\)

\(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

\(C\%_{HCl}=\dfrac{3,65.100}{111,7}=3,27\%\)

\(C\%_{MgCl_2}=\dfrac{9,5.100}{111,7}=8,5\%\)

Bài 10

a) \(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\); \(n_{H_2SO_4}=0,2.2,5=0,5\left(mol\right)\)

b) 

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,5}{3}\) => Fe₂O₃ hết, H₂SO₄ dư

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

              0,15------>0,45-------->0,15

=> \(n_{H_2SO_4\left(dư\right)}=0,5-0,45=0,05\left(mol\right)\)

c) \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=\dfrac{0,05}{0,2}=0,25M\\C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)

Bài 10:

a)

\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

\(n_{H_2SO_4}=0,2.2,5=0,5\left(mol\right)\)

b)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

bđ         0,15        0,5

pư        0,15----->0,45-------->0,15

spư        0           0,05            0,15

c)
\(\left\{{}\begin{matrix}C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,05}{0,2}=0,25M\\C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)

Bài 7:

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

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: n_Al(pư) = \(\dfrac{2}{3}.n_{H_2}=0,1\left(mol\right)\)

=> m_Al(pư) = 0,1.27 = 2,7 (g)

c) Theo PTHH: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2}=0,05\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

m_Al(dư) = 4,05 - 2,7 = 1,35 (g)

d) Theo PTHH:  \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)

=> \(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

...Hôm nọ bạn đăng bài rồi mà :) ?

\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\); \(n_{HNO_3}=\dfrac{100.18,9\%}{63}=0,3\left(mol\right)\)

PTHH: \(MgO+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}=\dfrac{0,3}{2}\) => Phản ứng vừa đủ

\(n_{Mg\left(NO_3\right)_2}=0,15\left(mol\right)\)

=> \(m_{Mg\left(NO_3\right)_2}=0,15.148=22,2\left(g\right)\)

\(C\%_{Mg\left(NO_3\right)_2}=\dfrac{22,2}{6+100}.100\%=20,94\%\)

- Hoà tan hỗn hợp vào nước dư, khuấy đều, sau đó lọc lấy chất rắn không tan là Fe₂O₃. NaCl và BaO tan hết trong nước.

`BaO + H_2O -> Ba(OH)_2`