Mg + 2HCl -> MgCl2 + H2 (1)
MgO + 2HCl -> MgCl2 + H2O (2)
nH2=0,2(mol)
Từ 1:
nMg=nH2=0,2(mol)
mMg=24.0,2=4,8(g)
mMgO=8,8-4,8=4(g)
a) PTHH: Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT1: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2\times24=4,8\left(g\right)\)
\(\Rightarrow m_{MgO}=8,8-4,8=4\left(g\right)\)