Gọi \(\left\{{}\begin{matrix}n_R=x\left(mol\right)\\n_{RO}=y\left(mol\right)\\n_{R\left(OH\right)_2}=z\left(mol\right)\\n_{RCO_3}=t\left(mol\right)\end{matrix}\right.\)
=> \(xM_R+y\left(M_R+16\right)+z\left(M_R+34\right)+t\left(M_R+60\right)=39\)
=> \(\left(x+y+z+t\right)M_R+16y+34z+60t=39\) (1)
PTHH: \(R+2HCl\rightarrow RCl_2+H_2\)
x--->2x------->x-------->x
\(RO+2HCl\rightarrow RCl_2+H_2O\)
y------>2y------->y
\(R\left(OH\right)_2+2HCl\rightarrow RCl_2+2H_2O\)
z--------->2z--------->z
\(RCO_3+2HCl\rightarrow RCl_2+CO_2+H_2O\)
t------->2t-------->t--------->t
=> 2x + 44t = 11,2 (2)
=> \(m_{ddHCl}=\dfrac{36,5.\left(2x+2y+2z+2t\right)}{14,6\%}=500\left(x+y+z+t\right)\left(g\right)\)
=> \(m_{ddspư}=500\left(x+y+z+t\right)+39-11,2=500\left(x+y+z+t\right)+27,8\left(g\right)\) (3)
=> \(C\%_{RCl_2}=\dfrac{\left(M_R+71\right)\left(x+y+z+t\right)}{500\left(x+y+z+t\right)+27,8}.100\%=17,5\%\) (4)
Ta có: \(n_{RCl_2}=0,65\left(mol\right)\)
=> x + y + z + t = 0,65 (5)
Thay (4), vào (5), ta được:
\(\dfrac{\left(M_R+71\right).0,65}{500.0,65+27,8}.100\%=17,5\%\)
=> MR = 24 (g/mol)
=> R là Mg
