n_H2SO4 = \(\frac{300.9,8\%}{98}\) = 0,3 (mol)

M₂O₃ + 3H₂SO₄ \(\rightarrow\) M₂(SO₄)₃ + 3H₂O

0,1 \(\leftarrow\) 0,3 ---------> 0,1 (mol)

M_M2O3 = \(\frac{10,2}{0,1}\)= 102 (g/mol)

\(\Rightarrow\) M = \(\frac{102-3.16}{2}\) = 27 (Al)

=> Al₂O₃

C%(muối)= \(\frac{0,1.342}{10,2+300}\) . 100% = 11,03 %

câu này dễ mà bn

Có đăng nhầm chỗ ko vậy bn???

CHÚC BẠN HỌC TỐT!!

Theo đề bài, ta có: \(\left\{{}\begin{matrix}n_{CO2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\n_{CaCl2}=\dfrac{66,6}{111}=0,6\left(mol\right)\end{matrix}\right.\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

pư............0,45............0,9............0,45..........0,45........0,45 (mol)

\(\Rightarrow n_{CaCl2\left(còn\right)}=0,6-0,45=0,15\left(mol\right)\)

PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2\)

pư..........0,15.........0,3...........0,15.........0,15 (mol)

a) % khối lượng mỗi chất trong A

\(m_{hhA}=m_{CaO}+m_{CaCO3}=56.0,15+100.0,45=53,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%CaO=\dfrac{56.0,15}{53,4}.100\%\approx15,73\%\\\%CaCO_3\approx84,27\%\end{matrix}\right.\)

b) \(\Rightarrow m_{ddHCl7,3\%}=\dfrac{36,5.\left(0,9+0,3\right)}{7,3\%}=600\left(g\right)\)

\(\Rightarrow V_{ddHCl7,3\%}=\dfrac{600}{1,1}\approx545,45\left(ml\right)\)

c) Ta có: \(m_{dds}=m_{hh}+m_{ddHCl}-m_{CO2}=53,4+600-44.0,45=633,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%CaCl_{2\left(I\right)}=\dfrac{111.0,15}{633,6}.100\%\approx2,63\%\\\%CaCl_{2\left(II\right)}=\dfrac{111.0,45}{633,6}.100\%\approx7,9\%\end{matrix}\right.\)