a) \(m_{ddHCl}=1,28.50=64\left(g\right)\)
Gọi \(\left\{{}\begin{matrix}n_{MgCO_3}=a\left(mol\right)\\n_{FeCO_3}=b\left(mol\right)\end{matrix}\right.\)
=> 84a + 116b = 8,32 (*)
\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)
\(n_{CO_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
PTHH:
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
a---------->2a------->a---------->a
\(FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\)
b---------->2b------>b---------->b
=> a + b = 0,08 (**)
=> a = 0,03; = 0,05
=> \(X:\left\{{}\begin{matrix}m_{MgCO_3}=0,03.84=2,52\left(g\right)\\m_{FeCO_3}=0,05.116=5,8\left(g\right)\end{matrix}\right.\)
b) Gọi \(n_{HCl\left(dư\right)}=c\left(mol\right)\)
Trong ddY có: \(\left\{{}\begin{matrix}MgCl_2:0,03\left(mol\right)\\FeCl_2:0,05\left(mol\right)\\HCl_{dư}:c\left(mol\right)\end{matrix}\right.\)
=> \(\dfrac{1}{2}\) ddY có: \(\left\{{}\begin{matrix}MgCl_2:0,015\left(mol\right)\\FeCl_2:0,0255\left(mol\right)\\HCl_{dư}:0,5c\left(mol\right)\end{matrix}\right.\)
PTHH:
\(MgCl_2+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2AgCl\downarrow\)
0,015----------------------------------->0,03
\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\)
0,025---------------------------------->0,05
\(HCl+AgNO_3\rightarrow HNO_3+AgCl\downarrow\)
0,5c--------------------------->0,5c
=> 0,5c + 0,05 + 0,03 = 0,1
=> c = 0,04
=> \(C\%_{HCl\left(bđ\right)}=\dfrac{\left(0,03.2+0,05.2+0,04\right).36,5}{64}.100\%=11,40625\%\)
c) \(m_{ddY}=8,32+64-0,08.44=68,8\left(g\right)\)
=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,03.95}{68,8}.100\%=4,14\%\\C\%_{FeCl_2}=\dfrac{0,05.127}{68,8}.100\%=9,23\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,04.36,5}{68,8}.100\%=2,12\%\end{matrix}\right.\)
