Mg + 2HCl \(\rightarrow\)MgCl2 + H2 (1)
MgO + 2HCl \(\rightarrow\)MgCl2 + H2O (2)
nH2=\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PTHH 1 ta có:
nMg=nH2=0,25(mol)
mMg=24.0,25=6(g)
mMgO=16-6=10(g)
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\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
MgO + 2HCl \(\rightarrow\) MgCl2 + H2O
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
de: 0,25 \(\leftarrow\) 0,25
\(m_{Mg}=0,25.24=6g\)
\(m_{MgO}=16-6=10g\)
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