\(m_{NaOH}=\dfrac{50.8}{100}=4\left(g\right)=>n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + H₂O

________0,1----->0,05

=> \(V=\dfrac{0,05}{2}=0,025\left(l\right)\)

\(n_{H_2SO_4}=0,5.0,05=0,025\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

_____0,05<---0,025

=> \(V_{ddKOH}=\dfrac{0,05}{2}=0,025\left(l\right)=25\left(ml\right)\)

=> C

CH3COOH + NaOH $\to$ CH3COONa + H2O

n NaOH = n CH3COOH = 50.6%/60 = 0,05(mol)

=> V dd NaOH = 0,05/2 = 0,025(lít) = 25(ml)

\(n_{CH_3COOH}=\dfrac{50\cdot6}{100\cdot60}=0.05\left(mol\right)\)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(0.05.................0.05\)

\(V_{NaOH}=\dfrac{0.05}{2}=0.025\left(l\right)\)

viết tách ra chứ vậy nản lắm

Theo đề ta có : nH2SO4 = 9,8/98 = 0,1(mol)

PTHH : ​

​2NaOH + h2SO4 - > Na2SO4 + H2O

​0,2mol......0,1mol..........0,2mol

​=> VNaOH = 0,2/2 = 0,1(l)

​=> mNa2SO4 = 0,1.142 = 14,2(g)

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\\ m_{H_2SO_4}=9,8\%.40=3,92\left(g\right)\\ n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

LTL: \(\dfrac{0,02}{2}< \dfrac{0,04}{3}\rightarrow\)H₂SO₄ dư

Theo pt: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,02=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,02=0,01\left(mol\right)\end{matrix}\right.\)

\(\rightarrow V_{H_2}=0,03.22,4=0,672\left(l\right)\\ m_{dd}=0,54+40=40,54\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,01}{40,54}=8,43\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,04-0,03\right).98}{40,54}=2,41\%\end{matrix}\right.\)

PTHH: HCl + NaOH --> NaCl + H₂O

______0,4<--0,4

=> m_NaOH = 0,4.40 = 16(g)

=> \(C\%NaOH=\dfrac{16}{80}.100\%=20\%\)