Question

hòa tan hết 2,04g oxit kim loại R hóa trị 3 cần dùng 500ml HCL 0,24M . Tìm ct oxit

Answer 1

R2O3+6HCl->2RCl3+3H2O

0,02--0,12 mol

n HCl=0,24.0,5=0,12 mol

=>M R2O3=2,04\0,02=102 đvC

=>R.2+16.3=102

=>R =27 (Al , nhôm )

=>Al2O3

Answer 2

\(R_2O_3+6HCl\rightarrow2RCl_3+3H_2O\\ n_{HCl}=0,5.0,24=0,12\left(mol\right)\\ n_{R_2O_3}=\dfrac{1}{6}.0,12=0,02\left(mol\right)\\ M_R=\dfrac{2,04}{0,02}=102\left(\dfrac{g}{mol}\right)\\ Mà:M_{R_2O_3}=2M_R+48\left(\dfrac{g}{mol}\right)\\ \Rightarrow2M_R+48=102\\ \Leftrightarrow M_R=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(III\right):Nhôm\left(Al=27\right)\\ CTHH.oxit:Al_2O_3\)