40. Although Mr. Brown was tired, he helped his wife with the housework.

=> Despite Mr Brown's tiredness, he helped his wife with the housework

\(n_{HCl}=0.5\cdot1=0.5\left(mol\right)\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.15.....0.3........................0.15\)

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

\(0.1.........0.5-0.3\)

\(m_A=0.15\cdot24+0.1\cdot80=11.6\left(g\right)\)

\(n_{CO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(n_{H_2O}=\dfrac{3.6}{18}=0.2\left(mol\right)\)

\(m_O=3-0.1\cdot12-0.2\cdot2=1.4\left(g\right)\)

X chứa : C , H , O 

CTHH là : \(R_xO_y\)

\(\%O=\dfrac{3}{7}\%R\)

\(\Rightarrow16y=\dfrac{3}{7}\cdot Rx\)

\(\Rightarrow\dfrac{112}{3}y=Rx\)

Với : \(x=2,y=3\Rightarrow R=56\)

\(Fe_2O_3\)

\(n_{AgCl}=\dfrac{43.05}{143.5}=0.3\left(mol\right)\) \(\Rightarrow n_{HCl}=0.3\left(mol\right)\)

\(n_{HCl}=\dfrac{6.72}{22.4}=0.3\left(mol\right),n_{Cl_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(H_2+Cl_2\underrightarrow{^{^{t^0}}}2HCl\)

\(0.15....0.15.......0.3\)

\(H\%=\dfrac{0.15}{0.2}\cdot100\%=75\%\)

\(n_{H_2SO_4}=n_{H_2}=\dfrac{2.8}{22.4}=0.125\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{kl}=18.8+0.125\cdot2-0.125\cdot98=6.8\left(g\right)\)

Bảo toàn khối lượng : 

\(m_{BaSO_4}=m_{BaCl_2}+m_{Na_2SO_4}-m_{NaCl}=20.8+14.2-11.8=23.2\left(g\right)\)

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{Br_2}=n_{C_2H_4}=\dfrac{2}{160}=0.0125\left(mol\right)\)

\(\Rightarrow n_{CH_4}=0.3-0.0125=0.2875\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(\%V_{CH_4}=\dfrac{0.2875}{0.3}\cdot100\%=95.83\%\)

\(\%V_{C_2H_4}=100\%-95.83\%=4.17\%\)

learning

\(n_{CaCO_3}=\dfrac{200}{100}=2\left(mol\right)\)

\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)

\(2.............................2\)

\(V_{CO_2}=2\cdot22.4=44.8\left(l\right)\)