nH2=$\frac{4,48}{22,4}$=0,2(mol)
a) PTHH: Mg+2CH3COOH→(CH3COO)2Mg+H2↑
0,2 0,2 0,2
MgO+2CH3COOH→(CH3COO)2Mg+H2O
0,4 0,4
b) mMg=0,2.24=4,8(g)=>mMgO=20,8-4,8=16 (g)
=>%mMg=$\frac{4,8}{20,8}$.100=23,08%
=>%mMgO=100%-23,08%=76,92%
c) nMgO=16/40=0,4(mol)
=>n(CH3COO)2Mg=0,2+0,4=0,6(mol)
=>m(CH3COO)2Mg=0,6.142=85,2(g)