Câu hỏi của Nguyễn Anh Khôi

Question

Cho 2,7gam bột nhôm tác dụ

Kudo Shinichi · Accepted Answer

$n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)$

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

0,1 0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

$n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)$

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: $0,4>0,15\rightarrow$ CuO dư

Theo pthh: $n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)$

$\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,15.64}{0,15.64+\left(0,4-0,15\right).80}=32,43\%\\%m_{CuO}=100\%-32,43\%=67,57\%\end{matrix}\right.$

Câu trả lời:

$n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)$

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

0,1 0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

$n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)$

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: $0,4>0,15\rightarrow$ CuO dư

Theo pthh: $n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)$

$\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,15.64}{0,15.64+\left(0,4-0,15\right).80}=32,43\%\\%m_{CuO}=100\%-32,43\%=67,57\%\end{matrix}\right.$

Nguyễn Ngọc Nam · Answer

a. $2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)$

b. $n_{Al}=\frac{m}{M}=\frac{2,7}{27}=0,1mol$

Theo phương trình `(1)` $n_{H_2}=\frac{3}{2}.n_{Al}=\frac{3}{2}.0,1=0,15mol$

$\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,15.22,4=3,36l$

c. $CuO+H_2\rightarrow^{t^o}Cu+H_2O\left(2\right)$

$n_{CuO}=\frac{m}{M}=\frac{32}{80}=0,4mol$

Tỷ lệ $\frac{0,4}{1}>\frac{0,15}{1}$

`->CuO` dư

Theo phương trình `(2)` $n_{Cu}=n_{H_2}=0,15mol$

$n_{CuO\left(pứ\right)}=n_{H_2}=0,15mol$

$\rightarrow n_{CuO\left(dư\right)}=0,4-0,15=0,25mol$

$m\left(g\right)\text{ chất rắn }\hept{\begin{cases}CuO_{dư}=0,25mol\Cu=0,15mol\end{cases}}$

$\rightarrow m=0,15.64+0,25.80=29,6g$

$\%m_{CuO\left(dư\right)}=\frac{0,25.80.100}{29,6}\approx67,6\%$

$\%m_{Cu}=100\%-67,6\%=32,4\%$