PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=n_{Na_2SO_4}=0,2mol\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,2\cdot98=19,6\left(g\right)\\m_{Na_2SO_4}=0,2\cdot142=28,4\left(g\right)\end{matrix}\right.\)

nNaOH=m/M=16:40=0.4(mol) 

pthh:2NaOH+H2SO4=Na2SO4+2H2O 

(mol)0,4         0,2         0,2   

mH2SO4=n.M=0,2.98=19,6(g)  

mNa2SO4=n.M=0,2.142=28,4(g)  

Ở pthh ko viết được mũi tên nên viết bằng dấu bằng nha

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

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

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

\(a)n_{H_2}=\dfrac{14,874}{24,79}=0,6mol\\2 Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,4mol     0,6mol          0,2mol           0,6mol

\(m_{Al}=0,4.27=10,8g\\ b)m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4g\\ c)m_{H_2SO_4}=0,6.98=58,8g\)

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)

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

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) 

PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1,8mol\\n_{FeCl_3}=0,6mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=1,8\cdot36,5=65,7\left(g\right)\\m_{FeCl_3}=0,6\cdot162,5=97,5\left(g\right)\end{matrix}\right.\)

a, PTHH: CuO + H2SO4 ----> CuSO4 + H2O

Ta có nCuO=32/80=0,4mol

Theo PTHH ta có: nH2SO4=nCuO=0,4mol

=> mH2SO4=0,4.98=39,2g

b, Theo PTHH ta có : nCuSO4=nCuO=0,4mol

=> mCuSO4=0,4.160=64g

Chúc bạn học tốt~

n_CuO = 0,4 mol

Pt: CuO + H₂SO₄ --> CuSO₄ + H₂O

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

m_H2SO4 = 0,4 . 98 = 39,2 (g)

m_CuSO4 = 0,4 . 160 = 64 (g)

PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O

Ta có: n_CuO= \(\dfrac{32}{80}\) = 0,4(mol)

Theo pthh, ta có: n_H2SO4 = n_CuO= 0,4(mol)

=> m_H2SO4= 0,4.98= 39,2(g)

b. Theo pthh, ta có: n_CuSO4= n_CuO= 0,4(mol)

=> mCuSO4= 0,4. 160= 64(g)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\\ n_{H_2SO_4}=n_{CuSO_4}=n_{CuO}=0,4\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,4.160=64\left(g\right)\\ m_{H_2SO_4}=98.0,4=39,2\left(g\right)\)

CuO+H₂SO₄->CuSO₄+H₂O

0,4------0,4---------0,4 mol

n _CuO=\(\dfrac{32}{80}\)=0,4 mol

=>m _CuSO4=0,4.160=64g

=>m _H2SO4=0,4.98=39,2g

\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

CuO + H₂SO₄ -> CuSO₄ + H₂O

  0,4        0,4           0,4

\(m_{CuSO_4}=0,4.160=64\left(g\right)\)

\(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

Mình nghĩ là vậy..