2,24ml chứ bn
\(n_A=\frac{V_A}{22,4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(M_A=\frac{m_A}{n_A}=\frac{1,7}{0,1}=17\left(g/mol\right)\)
\(CTTQ:X_xH_y\)
\(\%X=\frac{M_X.x.100}{M_A}\)
\(82,4=\frac{M_Xx.100}{17}\)
\(M_Xx=14\)
\(\%H=\frac{M_H.y.100}{M_A}\)
\(17,6=\frac{1.y.100}{17}\)
\(17,6=\frac{100y}{17}\)
\(y=3\)
\(Ta có\) \(M_A=M_X+M_H\)
\(\Leftrightarrow17=M_X+3\)
\(\Leftrightarrow M_X=14\)
\(\Rightarrow\) Có 1 nguyên tử X trong hợp chất
\(\Rightarrow X\) \(là \) \(nitơ\)
\(CTHH:NH_3\) ( lập cho cái CTHH :) )