$1)n_{O_2}=\dfrac{0,672}{22,4}=0,03(mol)$

Bảo toàn nguyên tố O: $n_O=2n_{O_2}=0,06(mol)$

$m_{O_2}=0,03.32=0,96(g)$

$\Rightarrow m_K+m_{Cl}=2,45-0,96=1,49(g)$

$\Rightarrow n_K=1,49.52,35\%\approx 0,02(mol)$

$\Rightarrow n_{Cl}=\dfrac{1,49-0,02.39}{35,5}=0,02(mol)$

Đặt CTPT của muối vô cơ là $K_xCl_yO_z$

$\Rightarrow x:y:z=0,02:0,02:0,06=1:1:3$

$\Rightarrow CTPT$ có dạng $(KClO_3)_n$ 

Mà $n=1\Rightarrow CTHH$ muối là $KClO_3$

$2)$

$a)PTHH:R_xO_y+2yHCl\to xRCl_{2y/x}+yH_2O$

Theo PT: $n_{R_xO_y}.2y=n_{HCl}=0,3$

$\Rightarrow \dfrac{12y}{x.M_R+16y}=0,15$

$\Rightarrow 12y=0,15.x.M_R+2,4y$

$\Rightarrow 0,15.x.M_R=9,6y$

$\Rightarrow M_R=\dfrac{64y}{x}$

Chọn $x=y=1\Rightarrow M_R=64(Cu)$

Vậy CT oxit là $CuO$

$b)PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$n_{CuO}=\dfrac{12}{80}=0,15(mol);n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

Xét tỉ lệ: $n_{CuO}>n_{H_2}\Rightarrow CuO$ dư

Theo PT: $n_{CuO}=n_{Cu}=n_{H_2}=0,1(mol)$

$\Rightarrow n_{CuO(dư)}=0,15-0,1=0,05(mol)$

$\Rightarrow m_{Cu(\text{thực tế})}=0,1.64.80\%=5,12(g)$

Và $m_{CuO(\text{thực tế})}=0,05.80.80\%=3,2(g)$

$\Rightarrow m_{rắn}=3,2+5,12=8,32(g)$