Question

Answer 1

CTHH: RHCO₃ (a mol), R₂CO₃ (b mol)

=> a.M_R + 61a + 2b.M_R + 60b = 13,45 (1)

\(n_{Ca\left(OH\right)_2}=1.0,075=0,075\left(mol\right)\)

PTHH: \(RHCO_3+HCl\rightarrow RCl+CO_2+H_2O\)

                   a------->a

             \(R_2CO_3+2HCl\rightarrow2RCl+CO_2+H_2O\)

                    b----->2b

              \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

                 0,075--->0,15

=> \(n_{HCl}=a+2b+0,15=0,3.1=0,3\left(mol\right)\)

=> a + 2b = 0,15 

(1) => 0,15.M_R + 31a + 4,5 = 13,45

=> 0,15.M_R + 31a = 8,95

Mà \(0< a< 0,15\)

=> 28,67 < M_R < 59,67

=> R là K

CTHH: KHCO₃, K₂CO₃

Ta có: \(\left\{{}\begin{matrix}100a+138b=13,45\\a+2b=0,15\end{matrix}\right.\)

=> a = 0,1 (mol); b = 0,025 (mol)

\(\left\{{}\begin{matrix}m_{KHCO_3}=0,1.100=10\left(g\right)\\m_{K_2CO_3}=0,025.138=3,45\left(g\right)\end{matrix}\right.\)