a) Ta có: \(2M_R+16=94\Rightarrow M_R=39\)
\(\Rightarrow R\) là K.
b) \(CTTQ:Fe_xO_y\)(x,y \(\in N\)*)
\(\%m_{Fe}=70\%\Rightarrow\dfrac{56x}{56x+16y}=0,7\)
\(\Rightarrow56x=39,2x+11,2\)
\(\Rightarrow16,8x=11,2y\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)
=> CTHH là Fe2O3
c) CTTQ: \(C_xO_y\)(x,y \(\in N\)*)
\(m_C:m_O=3:8\)
\(\Rightarrow12x:16y=\dfrac{12}{16}:\dfrac{y}{x}=\dfrac{3}{8}\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\)
=> CTHH là CO2 .
a.
\(2R+16=94\)
\(\Leftrightarrow R=\dfrac{94-16}{2}\)
\(\Leftrightarrow R=39\left(K\right)\)
\(\Rightarrow CTHH:K_2O\)