Đáp án:
$A$
Giải thích các bước giải:
Gọi số mol của $KHSO_4$ và $K_2CO_3$ lần lượt là $x, y$ mol
Vì TN2 có lượng khí thoát ra nhiều hơn TN1 nên ở thí nghiệm 1 vẫn muối $KHCO_3$.
TN2 còn muối $KHSO_4$ dư.
Xét TN1:
$KHS{O_4} + {K_2}C{O_3} \to KHC{O_3} + {K_2}S{O_4} $
$ y \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$ KHS{O_4} + KHC{O_3} \to {K_2}S{O_4} + C{O_2} + {H_2}O $
$ x - y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$ \to {n_{C{O_2}}} = x - y = \dfrac{{100 + 100 - 198,9}}{2} = 0,025\,\,(1) $
Xét TN2:
$2KHS{O_4} + {K_2}C{O_3} \to 2{K_2}S{O_4} + C{O_2} + {H_2}O $
$x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{x}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$ \to {n_{C{O_2}}} = \dfrac{{100 + 100 - 197,8}}{{44}} = \frac{x}{2} = 0,05\,\,mol $
$ \to x = 0,1 $
$ \to y = 0,075 $
$ \to {m_{KHS{O_4}}} = 0,1.136 = 13,6\,g$
$ \to C{\% _{KHS{O_4}}} = \dfrac{{13,6.100}}{{100}} = 13,6\% $
