Trong $27,75g$ quặng có $a$ mol $KCl$, $b$ mol $MgCl_2$, $c$ mol $H_2O$
$\Rightarrow 74,5a+95b+18c=27,75$ (1)
$KCl+AgNO_3\to KNO_3+AgCl$
$MgCl_2+2AgNO_3\to 2AgCl+Mg(NO_3)_2$
$n_{AgCl}=\dfrac{43,05}{143,5}=0,3(mol)$
$\Rightarrow a+2b=0,3$ (2)
Trong $55,5g=2.27,75g$ quặng có $2b$ mol $MgCl_2$, $2a$ mol $KCl$, $2c$ mol $H_2O$
$MgCl_2+2NaOH\to Mg(OH)_2+2NaCl$
$\Rightarrow 58.2b=11,6$ (3)
Từ $(1)(2)(3)\Rightarrow a=0,1; b=0,1; c=0,6$
$a: b: c=0,1:0,1:0,6=1:1:6$
$\Rightarrow x=y=1, t=6$
Vậy quặng là $KCl.MgCl_2.6H_2O$ (cacnalit)
CTPT: $KMgCl_3O_6H_{12}$