b

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{ZnCl_2}=0,3\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\\m_{ddHCl}=\dfrac{0,6\cdot36,5}{20\%}=109,5\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{19,5+109,5-0,6}\cdot100\%\approx37,78\%\)

1e

2a

3c

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{100}\cdot100\%=14,6\%\end{matrix}\right.\)

Chọn B

a) CO₂: 44 đvC

b) AgNO₃: 170 đvC

c) FeCl₃: 162,5 đvC

a) \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_O=4n_{Fe_3O_4}=0,08\left(mol\right)\end{matrix}\right.\)

b) \(n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_N=2n_{N_2O}=0,3\left(mol\right)\\n_O=n_{N_2O}=0,15\left(mol\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_H=2n_{H_2SO_4}=0,1\left(mol\right)\\n_S=n_{H_2SO_4}=0,05\left(mol\right)\\n_O=4n_{H_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\)

PTHH: \(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\uparrow\)

Ta có: \(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)=n_{KOH}\)

\(\Rightarrow C\%_{KOH}=\dfrac{0,2\cdot56}{7,8+200-0,1\cdot2}\cdot100\%\approx5,39\%\)

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)

           \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Na}=0,6\left(mol\right)\)

\(\Rightarrow\%m_{Na}=\dfrac{0,6\cdot23}{26,2}\cdot100\%\approx52,67\left(g\right)\) \(\Rightarrow\%m_{Na_2O}=47,33\%\)

Mặt khác: \(n_{Na_2O}=\dfrac{26,2-0,6\cdot23}{62}=0,2\left(mol\right)\)

Theo PTHH: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=1\left(mol\right)\) \(\Rightarrow m_{NaOH}=1\cdot40=40\left(g\right)\)