2Na + 2H2O -> 2NaOH + H2 (1)
2NaOH + MgSO4 -> Mg(OH)2 + Na2SO4 (2)
Mg(OH)2 -> MgO + H2O (3)
nNa=0,15(mol)
nMgSO4=0,1(mol)
Theo PTHH 1 ta có:
nNa=nNaOH=0,15(mol)
Vì \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\)nên MgSO4 dư 0,025 mol
Theo PTHH 2 ta có:
nMg(OH)2=\(\dfrac{1}{2}\)nNaOH=0,075(mol)
Theo PTHH 3 ta có:
nMg(OH)2=nMgO=0,075(mol)
mMgO=0,075.40=3(g)
nNa=\(\dfrac{3,45}{23}=0,15mol\)
\(n_{MgSO_4}=\dfrac{200.6}{120.100}=0,1mol\)
2Na+2H2O\(\rightarrow\)2NaOH+H2
\(n_{NaOH}=n_{Na}=0,15mol\)
MgSO4+2NaOH\(\rightarrow\)Mg(OH)2\(\downarrow\)+Na2SO4
-Tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{2}\rightarrow\)NaOH dư
\(n_{Mg\left(OH\right)_2}=n_{MgSO_4}=0,1mol\)
Mg(OH)2\(\rightarrow\)MgO+H2O
\(n_{MgO}=n_{Mg\left(OH\right)_2}=0,1mol\)
\(m_{MgO}=0,1.40=4gam\)
