Câu 1 :
\(CT:Fe_xO_y\)
\(\%O=\dfrac{16y}{160}\cdot100\%=30\%\)
\(\Leftrightarrow y=3\)
\(M=56x+16\cdot3=160\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow x=2\)
\(CT:Fe_2O_3\)
Câu 2 :
\(a.\)
\(CT:Ca_xC_yO_z\)
\(\text{Ta có :}\)
\(40x:12y:16z=3.33:1:4\)
\(\Leftrightarrow x:y:z=1:1:3\)
\(CTnguyên:\left(CaCO_3\right)_n\)
\(M_X=100\left(\dfrac{g}{mol}\right)\)
\(\Leftrightarrow100n=100\)
\(\Leftrightarrow n=1\)
\(CT:CaCO_3\)
Câu 2 :
\(b.\)
\(CT:H_xS_yO_z\)
\(x:y:z=\dfrac{2.04}{1}:\dfrac{32.65}{32}:\dfrac{65.31}{16}=2.04:1.02:4.08\)
\(\Leftrightarrow x:y:z=2:1:4\)
\(\text{CT nguyên : }\)\(\left(H_2SO_4\right)_n\)
\(M_Y=98\left(\dfrac{g}{mol}\right)\)
\(\Leftrightarrow98n=98\)
\(\Leftrightarrow n=1\)
\(CT:H_2SO_4\)
Câu 3 :
\(CT:N_xH_y\)
\(x:y=\dfrac{82.35}{14}:\dfrac{17.85}{1}=5.88:17.85=1:3\)
\(\text{CT nguyên : }\)\(\left(NH_3\right)_n\)
\(M_X=16\cdot1.0625=17\left(\dfrac{g}{mol}\right)\)
\(\Leftrightarrow17n=17\)
\(\Leftrightarrow n=1\)
\(CT:NH_3\)
