\(n_{Mg}=\dfrac{10}{24}=\dfrac{5}{12}\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{200\cdot9,125\%}{100\%}}{36,5}=0,5\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
n(bd) 5/12 0.5
n(spu) 1/6 0 0.25 0.25
\(m_{dd\left(spu\right)}=10+200-0,25\cdot2=209,5\left(g\right)\\ C\%_{MgCl_2}=\dfrac{0,25\cdot95}{209,5}\cdot100\%\approx11,337\%\)