CH3COOH+NaOH->CH3COONa+H2O

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

n CH3COOH=0,4 mol

=>CM NaOH=\(\dfrac{0,4}{0,1}=4M\)

\(n_{BaCl_2}=0,1.1=0,1\left(mol\right)\\ BaCl_2+K_2CO_3\rightarrow BaCO_3\downarrow+2KCl\\ Chất.tan.dd.sau.p.ứ:KCl\\ n_{KCl}=2.0,1=0,2\left(mol\right)\\ V_{ddsau}=V_{ddBaCl_2}+V_{ddK_2CO_3}=100+100=200\left(ml\right)=0,2\left(l\right)\\ \Rightarrow C_{MddKCl}=\dfrac{0,2}{0,2}=1\left(M\right)\)

n_NaOH = 0,1.2 = 0,2 (mol)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                0,2<--------0,2

=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

nNaOH = 0,1 . 2 = 0,2 (mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

nCH3COOH = nNaOH = 0,2 (mol)

CM(CH3COOH) = 0,2/0,15 = 1,33M

Bài 8:

nH2SO4=0,5(mool)

PTHH: 2 KOH + H2SO4 -> K2SO4 + 2 H2O

nKOH= 2.0,5=1(mol) => mKOH=1.56=56(g)

=> mddKOH= (56.100)/25=224(g)

Bài 7: 

mddNaOH= 2.1000.1,15=2300(g)

=> mNaOH=2300.30%=690(g)

=>nNaOH=690/40=17,25(mol)

??? Ủa xút là NaOH mà??

Bài 6:

nSO2= 0,784/22,4=0,035(mol)

PTHH: SO2 + Ca(OH)2 -> CaSO3 + H2O

nCa(OH)2=nSO2=0,035(mol)

=> CMddCa(OH)2= 0,035/0,25=0,14(M)

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PT: \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{KOH}=0,15\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na₂CO₃ dư.

Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)

Đáp án cần chọn là: C

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

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

\(n_{H_2SO_4}=0.1\cdot2=0.2\left(mol\right)\)

\(m_{dd_{H_2SO_4}}=100\cdot1.2=120\left(g\right)\)

\(n_{BaCl_2}=0.1\cdot1=0.1\left(mol\right)\)

\(m_{dd_{BaCl_2}}=100\cdot1.32=132\left(g\right)\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

\(0.1................0.1.........0.1...............0.2\)

\(\Rightarrow H_2SO_4dư\)

\(m_{BaSO_4}=0.1\cdot233=23.3\left(g\right)\)

\(V_{dd}=0.1+0.1=0.2\left(l\right)\)

\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2-0.1}{0.2}=0.5\left(M\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

\(m_{\text{dung dịch sau phản ứng}}=120+132-23.3=228.7\left(g\right)\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0.1\cdot98}{228.7}\cdot100\%=4.28\%\)

\(C\%_{HCl}=\dfrac{0.2\cdot36.5}{228.7}\cdot100\%=3.2\%\)

Giúp mình với 

=> Chọn đáp án B.