10.
$2R+2nHCl\to 2RCl_n+nH_2$
$n_R=\dfrac{6}{R}(mol)$
$\Rightarrow n_{RCl_n}=\dfrac{6}{R}(mol)$, $n_{H_2}=\dfrac{3n}{R}(mol)$
$m_{dd\text{spứ}}=6+100-\dfrac{2.3n}{R}=106-\dfrac{6n}{R}(g)$
$m_{RCl_n}=\dfrac{6(R+35,5n)}{R}=\dfrac{6R+213n}{R}(g)$
$\Rightarrow \dfrac{6R+213}{R}=0,2251(106-\dfrac{6n}{R})$
$\Leftrightarrow 6R+213=23,8606R-1,3506Rn$
$\Leftrightarrow R=\dfrac{213}{17,8606-1,3506n}$ (không có n thoả mãn R)