a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, Ta có: \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\)

\(\Rightarrow n_{CH_4}=\dfrac{11,2}{22,4}-0,2=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,3.16}{0,3.16+0,2.28}.100\%\approx46,15\%\\\%m_{C_2H_4}\approx53,85\%\end{matrix}\right.\)

c, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,7\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

0,7__________________0,7 (mol)

⇒ m_CaCO3 = 0,7.100 = 70 (g)

a, \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=0,1.3=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{3}\), ta được Al₂O₃ dư.

Theo PT: \(n_{Al_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow n_{Al_2O_3\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3\left(dư\right)}=0,1.102=10,2\left(g\right)\)

b, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

b, \(m_{NaOH}=32.25\%=8\left(g\right)\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

Theo PT: \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{32+24,8}.100\%=25\%\)

Ta có: \(C\%_{NaOH}=\dfrac{46,5}{200}.100\%=23,25\%\)

\(C_2H_4+H_2O\underrightarrow{^{xt,t^o}}C_2H_5OH\)

\(C_2H_5OH+O_2\underrightarrow{^{mengiam,t^o}}CH_3COOH+H_2O\)

Bạn xem lại đề chất cuối cùng nhé.

\(n_{Ag}=\dfrac{4,32}{108}=0,04\left(mol\right)\)

TH1: Hỗn hợp không có HCHO.

⇒ n_CnH2nO = 0,02 (mol)

\(\Rightarrow M_{C_nH_{2n}O}=\dfrac{1,02}{0,02}=51\left(g/mol\right)\)

⇒ n = 2,5

Mà: 2 aldehyde là đồng đẳng kế tiếp.

→ C₂H₄O và C₃H₆O.

CTCT: CH₃CHO và CH₃CH₂CHO

TH2: Hỗn hợp có HCHO.

Mà: 2 aldehyde đồng đẳng kế tiếp

→ HCHO và CH₃CHO

\(\Rightarrow\left\{{}\begin{matrix}30n_{HCHO}+44n_{CH_3CHO}=1,02\\4n_{HCHO}+2n_{CH_3CHO}=0,04\end{matrix}\right.\) → số mol âm (vô lý) → loại.

Ta có: n_NaOH = 0,2.1 = 0,2 (mol)

PT: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_{2\downarrow}\)

_______0,2______0,1_________________0,1 (mol)

a, \(V_{ddCuSO_4}=\dfrac{0,1}{2}=0,05\left(M\right)\)

b, \(m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

a, \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

b, Gọi n_{Fe pư} = x (mol)

Theo PT: n_Cu = n_Fe = x (mol)

Ta có: m _tăng  = m_Cu - m_Fe = 2,4

⇒ 64x - 56x = 2,4 ⇒ x = 0,3 (mol)

⇒ m_Fe = 0,3.56 = 16,8 (g)

m_Cu = 0,3.64 = 19,2 (g)

Ta có: \(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

___________0,2_____________0,2 (mol)

\(\Rightarrow m_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)

\(n_{C_2H_4}+n_{C_2H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(1\right)\)

Ta có: \(n_{Br_2}=\dfrac{40}{160}=0,25\left(mol\right)\)

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PT: n_C2H4 + 2n_C2H2 = n_Br2 = 0,25 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,15}{0,2}.100\%=75\%\\\%V_{C_2H_2}=25\%\end{matrix}\right.\)