Question

Answer 1

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1(mol)\\ n_{MgO}=\dfrac{4}{40}=0,1(mol)\\ n_{NaOH}=0,5.0,4=0,2(mol)\\ PTHH:Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ H_2SO_4+2NaOH\to Na_2SO_4+2H_2O\\ \Rightarrow \Sigma n_{H_2SO_4}=3n_{Al_2O_3}+n_{MgO}+\dfrac{1}{2}n_{NaOH}=0,1+0,3+0,1=0,5(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,5.98}{245}.100\%=20\%\)

Answer 2

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{NaOH}=0,4.0,5=0,2\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

_______0,1----->0,3__________________________(mol)

MgO + H₂SO₄ --> MgSO4 + H₂O

_0,1------>0,1________________________________(mol)

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

_0,2------->0,1________________________________(mol)

=> n_H2SO4 = 0,3 + 0,1 + 0,1 = 0,5 (mol)

=> m_H2SO4 = 0,5.98 = 49 (g)

=> \(C\%=\dfrac{49}{245}.100\%=20\%\)