Bàu 2

\(n_P=\dfrac{1,55}{31}=0,05mol\\ a)4P+5O_2\xrightarrow[]{t^0}2P_2O_5\)

0,05        0,0625       0,025

\(b)m_{P_2O_5}=0,025.142=3,55g\\ c)V_{O_2}=0,0625.24,79=1,549375l\)

Bài 3

\(a)n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2      0,4             0,2           0,2

\(V=V_{H_2}=0,2.24,79=4,958l\\ b)m_{FeCl_2}=0,2.127=25,4g\\ c)m_{ddHCl}=\dfrac{0,4.36,5}{20\%}\cdot100\%=73g\\ d)m_{dd}=11,2+73-0,2.2=83,8g\\ C_{\%FeCl_2}=\dfrac{25,4}{83,8}\cdot100\%=30,31\%\)

\(n_{Br_2}=\dfrac{640.10}{100.160}=0,4mol\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,4mol\\ \%V_{C_2H_4}=\dfrac{0,4.22,4}{12,1}\cdot100\%=74,05\%\\ \%V_{CH_4}=100-75,05=24,95\%\)

\(a)M_X=3,75.16=60g/mol\\ m_C=12\cdot\dfrac{4,48}{22,4}=2,4g\\ m_H=2\cdot\dfrac{3,6}{18}=0,4g\\ m_H+m_C=2,4+0,4=2,8g< m_X\\ m_O=6-2,8=3,2g\)

Trong X có C H O

Gọi CTPT X \(C_xH_yO_z\)

Ta có

\(\dfrac{12x}{2,4}=\dfrac{y}{0,4}=\dfrac{16z}{3,2}=\dfrac{60}{6}\\ \Rightarrow x=2;y=4;z=2\)ư

Vậy CTPT X \(CH_2COOH\)

\(b)n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,2mol\\ a=m_{CaCO_3}=0,2.100=20g\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1mol\\ V_{H_2,đktc}=0,1.22,4=2,24l=2440ml\\ V_{H_2,đkc}=0,1.24,79=2,479l=24790ml\)

mol Fe= 5,6:56=0,1(mol)
Fe + 2HCl => FeCl2 + H2
cứ 1mol Fe sẽ ra 1 mol H2
suy ra mol H2=0,1(mol)
suy ra  V H2= mol.22,4= 0,1.22,4= 2,24 lít= 2240 (ml)
 

\(n_{C_2H_4Br_2}=\dfrac{28,2}{188}=0,15mol\\ a)C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ b)n_{C_2H_4}=n_{C_2H_4Br_2}=0,15mol\\ \%V_{C_2H_4}=\dfrac{0,15.22,4}{5,6}\cdot100\%=60\%\\ \%V_{CH_4}=100\%-60\%=40\%\)

chọn A

A nhé

\(a)n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2mol\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{Br_2}=n_{C_2H_4}=n_{C_2H_4Br_2}=0,2mol\\ m_{Br_2}=0,2.160=32g\\ b)\%m_{C_2H_4}=\dfrac{0,2.28}{17,6}\cdot100\%=31,82\%\\ \%m_{CH_4}=100\%-31,82\%=68,18\%\)