a, PTHH: 2NaOH+ H2SO4--> Na2SO4+2H2O

ta có nNaOH=16/40=0,4mol

Theo PTHH ta có: nH2SO4=nNaOH.1/2=0,2mol

=> mH2SO4=0,2.98=19,6g

b, Theo PTHH ta có : nNa2SO4=nH2SO4=0,2mol

=> mNa2SO4=0,2.142=28,4g

Chúc bạn học tốt~

nNaOH = 0,4 mol

Pt: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

0,4 mol----> 0,2 mol---> 0,2 mol

m_H2SO4 = 0,2 . 98 = 19,6 (g)

m_Na2SO4 = 0,2 . 142 = 28,4 (g)

PTHH: \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15mol\\n_{Al_2\left(SO_4\right)_3}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\end{matrix}\right.\)

\(n_{Fe_2O_3} = \dfrac{48}{160} = 0,3(mol)\)

Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

0,3...........1,8..........0,6..............................(mol)

Vậy :

\(m_{HCl} = 1,8.36,5 = 65,7(gam)\\ m_{FeCl_3} = 0,6.162,5 = 97,5(gam)\)

nFe2O3 = 48/160 = 0.3 (mol) 

Fe2O3 + 6HCl => 2FeCl3 + 3H2O 

0.3_______1.8_____0.6 

mHCl = 1.8 * 36.5 = 65.7 (g) 

mFeCl3 = 0.6*162.5=97.5 (g) 

n_SO2 = 3.36/22.4 = 0.15 (mol) 

2Fe + 6H₂SO_4(đ) => Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

0.1...............................0.05.............0.15

m_Fe2O3 = 21.6 - 0.1*56 = 16 (g) 

n_Fe2O3 = 16/160 = 0.1 (mol) 

Fe₂O₃ + 3H₂SO₄ => Fe₂(SO₄)₃ + 3H₂O

0.1...................................0.1

m_Fe2(SO4)3 = ( 0.1 + 0.05) * 400 = 60 (g)

Bài 1:

\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot0,15=0,075\left(mol\right)\\ \Rightarrow m=m_{H_2SO_4}=0,075\cdot98=7,35\left(g\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V=V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)

$n_{Al_2O_3} = \dfrac{4,08}{102} = 0,04(mol)$

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,04          0,12                    0,04                                (mol)

$m_{Al_2(SO_4)_3} = 0,04.342 = 13,68(gam)$

$C\%_{H_2SO_4} = \dfrac{0,12.98}{280}.100\% = 4,2\%$

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,09\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,09\cdot98}{9,8\%}=90\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,03\cdot400=12\left(g\right)\end{matrix}\right.\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

a

PTPỨ: 

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,1------>0,1------>0,1

b 

\(m_{muối}=m_{CuSO_4}=0,1.160=16\left(g\right)\)

\(V_{dd.H_2SO_4.pứ}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)

Học kỹ công thức rồi hãy trả lời câu hỏi nhé ctv kì 21: )

\(CM=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{CM}\)