PTPƯ:
Mg + 2HCl -> MgCl2 + H2 (1)
.x........2x..........x...........x
MgO + 2HCl -> MgCl2 + H2O (2)
.y...........2y...........y...........y
nMgCl2 = \(\dfrac{28,5}{95}=0,3\) mol
Từ (1) và (2) ta có hpt: \(\left\{{}\begin{matrix}24x+40y=8,8\\x+y=0,3\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
mMg = 0.2.24 = 4,8 g
a) C% Mg = \(\dfrac{4,8}{8,8}\).100% = 54,55%
C%MgO = 100% - 54,55% = 45,45%
b) mHCl = (0,4+0,2).36,5 = 21,9 g
mdd HCl = \(\dfrac{21,9.100}{14,6}\)= 150 g
c) mdd sau pư = 8,8 + 150 = 158,8 g
C% MgCl2 = \(\dfrac{28,5}{158,8}100\%\) = 17,9 %