pt CH3COOH+NaOH\(\rightarrow\)CH3COONa+H2O

ta có mCH3COOH=15*120:100=18 g\(\Rightarrow\)nCH3COOH=0,3 mol

mNaOH=20*100:100=20g\(\Rightarrow\)nNaOH=0,5 mol

theo pt thì NaOH dư

ta có nCH3COONa=nCH3COOH=0,3 mol\(\Rightarrow\)mCH3COONa=24,6 g

ta có m dd sau phản ứng =120+100=220 g \(\Rightarrow\)C%CH3COONa=11,2%

a) \(m_{NaOH}=60.5\%=3\left(g\right)\)

\(m_{dd}=40+60=100\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{3}{100}.100\%=3\%\)

b) \(m_{dd}=100+50=150\left(g\right)\)

\(m_{KOH}=100.20\%+50.15\%=27,5\left(g\right)\)

\(\Rightarrow C\%\left(KOH\right)=\dfrac{27,5}{150}.100\%=18,33\%\)

a, m_ddsau = 40 + 60 = 100 ( g )

m_NaOH = 3 ( g )

=> \(C\%=\dfrac{3}{100}.100=3\%\)

b, - Gọi nồng độ dd thu được là X % .

- Áp đụng pp đường chéo ta có :

\(\Rightarrow\dfrac{100}{50}=2=\dfrac{X-15}{20-X}\)

=> X = 55/3 %

\(n_{HCl}=\dfrac{400.18,25\%}{36,5}=2\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,25.2=0,5\left(mol\right)\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ 0,5...............1...............0,5\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ 1............1..............1\left(mol\right)\\ C\%_{ddNaOH}=\dfrac{1.40}{120}.100\approx33,333\%\)

Câu 8 : 

\(n_{MgCO3}=\dfrac{42}{84}=0,5\left(mol\right)\)

Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

                 1                  0,5                                         0,5

a) \(V_{CO2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

b) \(V_{CH3COOH}=\dfrac{1}{2}=0,5\left(l\right)\)

c) Pt : \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

                  1                                         1

300ml = 0,3l

\(C_{MCH3COONa}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)

 Chúc bạn học tốt

\(C\%_{ddNaOH\left(thu.được\right)}=\dfrac{20}{20+150}.100\%\approx11,765\%\)

Câu 1:

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\) 

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

            \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{12,4}{62}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4\cdot40}{12,4+193,8}\cdot100\%\approx7,76\%\)

b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{CuSO_4}=\dfrac{100\cdot16\%}{160}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,4}{2}>\dfrac{0,1}{1}\) \(\Rightarrow\) NaOH còn dư, CuSO₄ p/ứ hết

\(\Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)

Bài 2 : 

a)

$Cu + 2H_2SO_{4_{đặc}} \to CuSO_4 + SO_2 + 2H_2O$
$n_{Cu} = n_{SO_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$\%m_{Cu} = \dfrac{0,1.64}{15}.100\% = 42,67\%$
$\%m_{CuO} = 100\% -42,67\% = 57,33\%$

b)

$NaOH + SO_2 \to NaHSO_3$
$n_{NaOH} = n_{SO_2} = 0,1(mol)$

$\Rightarrow V_{dd\ NaOH} = \dfrac{0,1}{1} = 0,1(lít) = 100(ml)$

pt 2 CH3COOH+Mg\(\rightarrow\)(CH3COO)2Mg+H2

nMg=7,2:24=0,3 mol

mCH3COOH=120*20:100=24g suy ra nCH3COOH=0,4 mol

theo pt thì Mg dư

theo pt n(CH3COO)2Mg=1/2nCH3COOH=0,2 mol suy ra n(CH3COO)2Mg=28,4g

mdd sau phản ứng =7,2+120-0,4=126,8 g

suy ra C%(CH3COO)Mg=22,3%