Ta có: \(n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\)
PTHH: MgO + 2HCl ---> MgCl2 + H2O
Ta có: \(m_{dd_{MgCl_2}}=10+115=125\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,25\left(mol\right)\)
=> \(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{23,75}{125}.100\%=19\%\)