PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=2,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=280\left(l\right)\)

Ta có: \(n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)=n_C\)

⇒ m_H = 4,4 - m_C = 0,8 (mol) ⇒ n_H = 0,8 (mol)

M_X = 22.2 = 44 (g/mol)

\(\Rightarrow n_X=\dfrac{4,4}{44}=0,1\left(mol\right)\)

Số nguyên tử C là: \(\dfrac{n_C}{n_X}=\dfrac{0,3}{0,1}=3\)

Số nguyên tử H là: \(\dfrac{n_H}{n_X}=\dfrac{0,8}{0,1}=8\)

→ CTPT của X là C₃H₈.

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(CH\equiv C-CH_3+2Br_2\rightarrow CHBr_2-CBr_2-CH_3\)

\(CH_2=CH-CH_3+Br_2\rightarrow CH_2Br-CHBr-CH_3\)

- Hợp chất vô cơ: NaHCO₃, CO₂, CaCO₃, Na₂CO₃.

- Hợp chất hữu cơ:

+ Hydrocacbon: CH₄, C₂H₄, C₂H₂, C₄H₁₀.

+ Dẫn xuất hydrocacbon: CH₃OH, C₃H₈O, CH₃Br, C₂H₆O.

\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)

tỉ lệ         4     :      5        :   2

n(mol)    0,2--->0,25-------->0,1

\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ V_{kk}=5,6:\dfrac{1}{5}=28\left(l\right)\)

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)

\(\Rightarrow V_{kk}=5V_{O_2}=28\left(l\right)\)

a, Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)

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

THeo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow V_{O_2}=0,04.22,4=0,896\left(l\right)\)

b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)

a) \(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02\left(mol\right)\).

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

Mol :       3    :     2   :     1

Mol                  0,04 ←  0,02

\(\Rightarrow V_{O_2}=n_{O_2}.22,4=\left(0,04\right).\left(22,4\right)=0,896\left(l\right)\).

b) Từ phương trình ở câu a \(\Rightarrow n_{O_2}=0,04\left(mol\right)\).

PTHH : \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Mol :           2         :        1         :       1     :   1

Mol :        0,08                    ←                     0,04

\(\Rightarrow m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\left(0,08\right).158=12,64\left(g\right)\).

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

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

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\), ta được H₂ dư.

Theo PT: \(n_{H_2\left(pư\right)}=n_{Cu}=n_{H_2O}=n_{CuO}=0,15\left(mol\right)\)

\(\Rightarrow n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\Rightarrow m_{H_2\left(dư\right)}=0,05.2=0,1\left(g\right)\)

\(m_{Cu}=0,15.64=9,6\left(g\right)\)

\(m_{H_2O}=0,15.18=2,7\left(g\right)\)

Bạn xem lời giải ở đây nhé.

https://hoc24.vn/cau-hoi/hoa-tan-108-g-mg-vao-dd-h2so4-20-vua-du-sau-khi-phan-ung-ket-thuc-thu-duoc-dd-x-lam-lanh-dd-x-xuong-20-do-c-thu-duoc-1476-g-muoi-sunfat-ket-ti.4797186776937