200ml = 0,2l

\(n_{KOH}=1.0,2=0,2\left(mol\right)\)

Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O|\)

            2              1               1            2

           0,2           0,1

\(n_{H2SO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{H2SO4}=0,1.98=9,8\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{9,8.100}{49}=20\)^0/₀

 Chúc bạn học tốt

a, \(n_{KOH}=0,3.1=0,3\left(mol\right)\)

PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Theo PT: \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{K_2SO_4}=0,15.174=26,1\left(g\right)\)

b, \(C_{M_{K_2SO_4}}=\dfrac{0,3}{0,3+0,2}=0,6\left(M\right)\)

\(a,n_{H_2SO_4}=1.0,1=0,1(mol)\\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{naOH}=2n_{H_2SO_4}=0,2(mol)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{0,2.40}{10\%}=80(g)\\ b,m_{dd_{H_2SO_4}}=1,2.100=120(g)\\ n_{Na_2SO_4}=0,1(mol)\\ \Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{80+120}.100\%=7,1\%\)

nK2SO3=15.8\158=0.1(mol)

K2SO3+H2SO4→K2SO4+SO2+H2O

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

VSO2=0.1⋅22.4=2.24(l)

CMK2SO4=0.1\0.2=0.5(M)

CuO+H₂SO₄->CuSO₄+H₂O

0,02----0,02 mol

n _CuO=\(\dfrac{1,6}{80}\)0,02 mol

=>Cm _H2SO4=\(\dfrac{0,02}{0,1}\)=0,2 M

Đổi 100 ml = 0,1 l

CuO + H₂SO₄ -> H₂O + CuSO₄

Ta có \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(=>Cm_{H_2SO_4}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Ta có: 40n_NaOH + 56n_KOH = 25,44 (1)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}+\dfrac{1}{2}n_{KOH}=0,3.0,9=0,27\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,3\left(mol\right)\\n_{KOH}=0,24\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3.40=12\left(g\right)\\m_{KOH}=0,24.56=13,44\left(g\right)\end{matrix}\right.\)

b, \(m_{ddH_2SO_4}=300.1,14=342\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,27.98}{342}.100\%\approx7,74\%\)

Câu 2 đề cho hết rồi kìa cậu

nZn = 13/65 = 0.2 (mol) 

Zn + H2SO4 => ZnSO4 + H2

0.2......0.2..........................0.2 

VH2 = 0.2*22.4 = 4.48 (l)

C%H2SO4 = 0.2*98/200 * 100% = 9.8 %

nCuO = 8/80 = 0.1 (mol) 

CuO + H2 -to-> Cu + H2O 

0.1......0.1...........0.1 

=> H2 dư 

mCu = 0.1*64 = 6.4 (g) 

â) nZn=0,2(mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

0,2_____0,2______0,2_____0,2(mol)

=> V(H2,đktc)=0,2.22,4=4,48(l)

b) C%ddH2SO4= [(98.0,2)/200)].100=9,8%

c) nCuO=0,1(mol)

PTHH: CuO + H2 -to-> Cu + H2O

Ta có: 0,1/1 < 0,2/1

=> H2 dư, CuO hết, tính theo nCuO

=> nCu=nCuO=0,1(mol)

=>mCu=6,4(g)