Gọi: \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\\n_{FeO}=c\left(mol\right)\end{matrix}\right.\)

- Khi cho pư với H₂.

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{CuO}+n_{FeO}=a+c=0,5\left(1\right)\)

\(\left\{{}\begin{matrix}n_{Cu}=n_{CuO}=a\left(mol\right)\\n_{Fe}=n_{FeO}=c\left(mol\right)\end{matrix}\right.\)

⇒ 64a + 56c = 29,6 (2)

- Cho hh pư với dd HCl.

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

Ta có: \(n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CuO}+2n_{MgO}+2n_{FeO}=2a+2b+2c=1,2\left(3\right)\)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\\c=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{MgO}=0,1.40=4\left(g\right)\\m_{FeO}=0,3.72=21,6\left(g\right)\end{matrix}\right.\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,6}{2}\), ta được HCl dư.

Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\)

c, Theo PT: \(n_{HCl\left(pư\right)}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,6-0,5=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

m_ZnCl2 = 34 (g) (theo phần b)

Bạn có thể lên mạng để tìm nhá!!

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

C1: B

C2: A

C3: C

C4: B

C5: B

C6: B

PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Ta có: \(n_R=\dfrac{9,6}{M_R}\left(mol\right)\)

\(n_{RO}=\dfrac{16}{M_R+16}\left(mol\right)\)

Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{9,6}{M_R}=\dfrac{16}{M_R+16}\Rightarrow M_R=24\left(g/mol\right)\)

→ R là Mg.

C8: Ta có: \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

____0,1____________0,1____0,05 (mol)

⇒ m_KOH = 0,1.56 = 5,6 (g)

V_H2 = 0,05.22,4 = 1,12 (l)

C9: A

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

C10: A

\(12x-33=3^2.3^3\Leftrightarrow\)\(3\left(4x-11\right)=3^5\Leftrightarrow4x-11=3^4\)

\(\Leftrightarrow x=\frac{35}{2}\)