$n_{HCl}=0,1 mol$
$n_{H_2}=\dfrac{1,68}{22,4}=0,075 mol$
$2K+2HCl\to 2KCl+H_2$ (1)
$n_{K(1)}=n_{KCl}=n_{HCl}=0,1 mol$
$n_{H_2(1)}=0,01:2=0,05 mol < 0,075$
$\Rightarrow$ Có phản ứng kali với nước.
$n_{H_2(2)}=0,075-0,05=0,025 mol$
$2K+2H_2O\to 2KOH+H_2$ (2)
$\Rightarrow n_{KOH}=2n_{H_2(2)}=0,025.2=0,05 mol$
$m_{\text{rắn}}= m_{KCl}+ m_{KOH}=0,1.74,5+0,05.56=10,25g$