\(n_{NaOH}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH
______0,2-------------->0,4

=> m_NaOH = 0,4.40 = 16(g)

=> \(C\%=\dfrac{16}{12,4+187,6}.100\%=8\%\)

=> B

46

\(m_{NaOH}=100\cdot4\%=4\left(g\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.05.......................0.1\)

\(m_{Na_2O}=0.05\cdot62=3.1\left(g\right)\)

a) Na2CO3 + 2 CH3COOH -> 2 CH3COONa + H2O + CO2

nNa2CO3= 0,1(mol)

b) mddCH3COOH= 0,95.200= 190(g)

c) => mCH3COOH= 12% . 190=22,8(g)

=> nCH3COOH= 0,38(mol)

Ta có: 0,1/ 1 < 0,38/2

=> Na2CO3 hết, CH3COOH dư, tính theo nNa2CO3.

- Chất trong dd sau phản ứng CH3COONa và CH3COOH dư.

mddsaup.ứ= mNa2CO3 + mddCH3COOH - mCO2= 10,6+190-0,1. 44= 196,2(g)

\(C\%ddCH3COONa=\frac{0,2.82}{196,2}.100\approx8,359\%\\ C\%ddCH3COOH\left(Dư\right)=\frac{\left(0,38-0,2\right).60}{196,2}.100\approx5,505\%\)

Vddsau=VddCH3COOH= 0,2(l)

\(C_{MddCH3COONa}=\frac{0,2}{0,2}=1\left(M\right)\\ C_{MddCH3COOH\left(dư\right)}=\frac{0,38-0,2}{0,2}=0,9\left(M\right)\)

Bài 2:

\(n_{Na2O}=\frac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1_______________0,2

\(n_{CuSO4}=\frac{300.16\%}{160}=0,3\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

Trước_____ 0,3_______0,2________________

Phản ứng__0,1______0,2____________

Sau_______ 0,2____ 0 _________ 0,1

\(Cu\left(OH\right)_2\rightarrow CuO+H_2O\)

0,1 __________0,1

\(\rightarrow m=0,1.80=8\left(g\right)\)

Bài 3:

\(n_{Na2CO3}=\frac{10,6}{160}=0,1\left(mol\right)\)

\(n_{H2SO4}=\frac{150.9,8\%}{98}=0,15\left(mol\right)\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(n_{H2SO4_{du}}=0,15-0,1=0,05\left(mol\right)\)

\(n_{CO2}=n_{Na2CO3}=0,1\left(mol\right)\)

\(m_{dd_{spu}}=10,6+150-0,144=156,2\left(g\right)\)

\(C\%_{H2SO4}=\frac{0,05.98}{156,2}.100\%=3,14\%\)

\(C\%_{Na2SO4}=\frac{0,1.142}{156,2}.100\%=9,09\%\)

Bài 4:

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\left(1\right)\)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\left(2\right)\)

Theo ptpu, ta có:

\(n_{Ba\left(OH\right)2\left(1\right)}=\frac{1}{2}n_{HCl}=0,03\left(mol\right)\)

\(n_{Ba\left(OH\right)2\left(2\right)}=n_{H2SO4}=0,02\left(mol\right)\)

\(n_{Ba\left(OH\right)2_{can}}=n_{Ba\left(OH\right)2\left(1\right)}+n_{Ba\left(OH\right)2\left(2\right)}=0,05\left(mol\right)\)

\(\rightarrow V_{Ba\left(OH\right)2_{Can}}=\frac{0,05}{0,2}=0,25l=250\left(ml\right)\)

Câu 1:

\(n_{K2O}=\frac{9,4}{39.2+16}=0,1\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

0,1_____________0,2

\(C\%_{KOH}=\frac{0,2.\left(39+17\right)}{150,6+9,4}.100\%=7\%\)

\(KOH+HCl\rightarrow KCl+H_2O\)

0,2______0,2__________________

\(\Rightarrow V_{dd_{HCl}}=\frac{0,2}{0,5}=0,5\left(l\right)\)

Câu 2:

a, \(n_{K2O}=\frac{23,5}{39.2+16}=0,25\left(mol\right)\)

\(2n_{K2O}=n_{KOH}\Rightarrow n_{KOH}=0,25.2=0,5\left(mol\right)\)

\(C\%_{KOH}=\frac{0,5.\left(39+17\right)}{176,5+23,5}.100\%=14\%\)

b, \(n_{KOH}=2n_{K2SO4}\Rightarrow n_{K2SO4}=\frac{0,5}{2}=0,25\)

\(n_{H2SO4}=n_{K2SO4}=0,25\)

\(m_{dd_{H2SO4}}=\frac{0,25.98}{20\%}=122,5\left(g\right)\)

c,

mdd sau phản ứng=mddA+mddH2SO4

m dd sau phản ứng \(=23,5+176,5+122,5=322,5\)

\(C\%_{K2SO4}=\frac{0,25.\left(39.2+32+16.4\right)}{322,5}.100\%=13,49\%\)

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