\(\dfrac{M_A}{M_{H2}}=18,25\)
=> MA = 36,5 g/mol
và %Cl = 97,26%
Đặt CTHH là HxCly
\(\dfrac{x}{2,74}=\dfrac{35,5y}{97,26}=\dfrac{36,5}{100}=0,365\)
x = \(\dfrac{0,365.2,74}{1}=1\)
y = \(\dfrac{0,365.97,26}{35,5}=1\)
Vậy CTHH cần lập là HCl
Ta có \(d_{\dfrac{A}{H_2}}=18,25\Rightarrow M_A=18,25.236,5\left(\dfrac{g}{mol}\right)\\ m_H=2,74\%.36,5=1\left(g\right)\Rightarrow n_H=\dfrac{1}{1}=1\left(mol\right)\\ \%Cl=100\%-2,74\%=97,26\%\\ m_{Cl}=97,26\%.36,5\approx35,5\left(g\right)\)
\(\Rightarrow n_{Cl}=\dfrac{35,5}{35,5}=1\left(mol\right)\\ \Rightarrow CTHH.là:HCl\)
a. \(M_A=18,25.2=36,5\left(g\right)\)
Gọi CTHH của A là: \(\left(H_xCl_y\right)_n\)
Ta có: \(x:y=\dfrac{2,74\%}{1}:\dfrac{100\%-2,74\%}{35,5}=2,74:2,74=1:1\)
Vậy CTHH của A là: \(\left(HCl\right)_n\)
Ta có: \(M_A=\left(1+35,5\right).n=36,5\left(g\right)\)
\(\Leftrightarrow n=1\)
Vậy A là khí hiđro clorua (HCl)
a) \(d_{\dfrac{A}{H_2}}=\dfrac{M_A}{2}=18,25\Rightarrow M_A=18,25.2=36.5\left(gmol\right)\)
b) \(m_H=\dfrac{36,5.2,74\%}{100\%}=1\left(g\right)\\ n_H=\dfrac{1}{1}=1\left(mol\right)\\ m_{Cl}=36,5-1=35,5\left(g\right)\\ n_{Cl}=\dfrac{35,5}{35,5}=1\left(mol\right)\\ =>CTHH:HCl\)
